Does a random sequence of vectors span a Hilbert space?












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Let $mathcal{H}$ be a separable Hilbert space. Let $v$ be a random variable taking values in $mathcal{H}$ such that $P(v perp h) < 1$ for all $h in mathcal{H}.$ Suppose we sample an infinite sequence $v_1, v_2, ldots.$ Is it the case that, almost surely, the closed span of $v_1, v_2, ldots$ is all of $mathcal{H}?$










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$endgroup$












  • $begingroup$
    and if they are iid the probability of being in a closed hyperplane $(h)^perp$ is $P(v_kperp h, k=1,2,dots)=0$
    $endgroup$
    – Pietro Majer
    Apr 21 at 17:07






  • 3




    $begingroup$
    @Pietro Majer: this is the probability that the vectors all lie in a given closed hyperplane.
    $endgroup$
    – Anthony Quas
    Apr 21 at 17:10








  • 3




    $begingroup$
    Whoever voted to close this, I'm pretty sure you don't understand the question. This is subtle and interesting.
    $endgroup$
    – Anthony Quas
    Apr 21 at 17:18






  • 1




    $begingroup$
    Maybe I'm missing something obvious, but is it even clear that the event you are interested in is measurable?
    $endgroup$
    – Jochen Glueck
    Apr 21 at 17:21








  • 1




    $begingroup$
    @JochenGlueck: Yes it's measurable: Let $(y_n)$ be a dense sequence in $mathcal H$. Then the event is: for all $m>0$, for all $n>0$, there exist $k>0$ and rational $t_1,ldots,t_k$ such that $|t_1v_1+ldots+t_kv_k-y_n|<1/m$.
    $endgroup$
    – Anthony Quas
    Apr 21 at 19:18


















8












$begingroup$


Let $mathcal{H}$ be a separable Hilbert space. Let $v$ be a random variable taking values in $mathcal{H}$ such that $P(v perp h) < 1$ for all $h in mathcal{H}.$ Suppose we sample an infinite sequence $v_1, v_2, ldots.$ Is it the case that, almost surely, the closed span of $v_1, v_2, ldots$ is all of $mathcal{H}?$










share|cite|improve this question









$endgroup$












  • $begingroup$
    and if they are iid the probability of being in a closed hyperplane $(h)^perp$ is $P(v_kperp h, k=1,2,dots)=0$
    $endgroup$
    – Pietro Majer
    Apr 21 at 17:07






  • 3




    $begingroup$
    @Pietro Majer: this is the probability that the vectors all lie in a given closed hyperplane.
    $endgroup$
    – Anthony Quas
    Apr 21 at 17:10








  • 3




    $begingroup$
    Whoever voted to close this, I'm pretty sure you don't understand the question. This is subtle and interesting.
    $endgroup$
    – Anthony Quas
    Apr 21 at 17:18






  • 1




    $begingroup$
    Maybe I'm missing something obvious, but is it even clear that the event you are interested in is measurable?
    $endgroup$
    – Jochen Glueck
    Apr 21 at 17:21








  • 1




    $begingroup$
    @JochenGlueck: Yes it's measurable: Let $(y_n)$ be a dense sequence in $mathcal H$. Then the event is: for all $m>0$, for all $n>0$, there exist $k>0$ and rational $t_1,ldots,t_k$ such that $|t_1v_1+ldots+t_kv_k-y_n|<1/m$.
    $endgroup$
    – Anthony Quas
    Apr 21 at 19:18
















8












8








8


3



$begingroup$


Let $mathcal{H}$ be a separable Hilbert space. Let $v$ be a random variable taking values in $mathcal{H}$ such that $P(v perp h) < 1$ for all $h in mathcal{H}.$ Suppose we sample an infinite sequence $v_1, v_2, ldots.$ Is it the case that, almost surely, the closed span of $v_1, v_2, ldots$ is all of $mathcal{H}?$










share|cite|improve this question









$endgroup$




Let $mathcal{H}$ be a separable Hilbert space. Let $v$ be a random variable taking values in $mathcal{H}$ such that $P(v perp h) < 1$ for all $h in mathcal{H}.$ Suppose we sample an infinite sequence $v_1, v_2, ldots.$ Is it the case that, almost surely, the closed span of $v_1, v_2, ldots$ is all of $mathcal{H}?$







reference-request fa.functional-analysis pr.probability operator-theory hilbert-spaces






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share|cite|improve this question










asked Apr 21 at 16:33









J. E. PascoeJ. E. Pascoe

575316




575316












  • $begingroup$
    and if they are iid the probability of being in a closed hyperplane $(h)^perp$ is $P(v_kperp h, k=1,2,dots)=0$
    $endgroup$
    – Pietro Majer
    Apr 21 at 17:07






  • 3




    $begingroup$
    @Pietro Majer: this is the probability that the vectors all lie in a given closed hyperplane.
    $endgroup$
    – Anthony Quas
    Apr 21 at 17:10








  • 3




    $begingroup$
    Whoever voted to close this, I'm pretty sure you don't understand the question. This is subtle and interesting.
    $endgroup$
    – Anthony Quas
    Apr 21 at 17:18






  • 1




    $begingroup$
    Maybe I'm missing something obvious, but is it even clear that the event you are interested in is measurable?
    $endgroup$
    – Jochen Glueck
    Apr 21 at 17:21








  • 1




    $begingroup$
    @JochenGlueck: Yes it's measurable: Let $(y_n)$ be a dense sequence in $mathcal H$. Then the event is: for all $m>0$, for all $n>0$, there exist $k>0$ and rational $t_1,ldots,t_k$ such that $|t_1v_1+ldots+t_kv_k-y_n|<1/m$.
    $endgroup$
    – Anthony Quas
    Apr 21 at 19:18




















  • $begingroup$
    and if they are iid the probability of being in a closed hyperplane $(h)^perp$ is $P(v_kperp h, k=1,2,dots)=0$
    $endgroup$
    – Pietro Majer
    Apr 21 at 17:07






  • 3




    $begingroup$
    @Pietro Majer: this is the probability that the vectors all lie in a given closed hyperplane.
    $endgroup$
    – Anthony Quas
    Apr 21 at 17:10








  • 3




    $begingroup$
    Whoever voted to close this, I'm pretty sure you don't understand the question. This is subtle and interesting.
    $endgroup$
    – Anthony Quas
    Apr 21 at 17:18






  • 1




    $begingroup$
    Maybe I'm missing something obvious, but is it even clear that the event you are interested in is measurable?
    $endgroup$
    – Jochen Glueck
    Apr 21 at 17:21








  • 1




    $begingroup$
    @JochenGlueck: Yes it's measurable: Let $(y_n)$ be a dense sequence in $mathcal H$. Then the event is: for all $m>0$, for all $n>0$, there exist $k>0$ and rational $t_1,ldots,t_k$ such that $|t_1v_1+ldots+t_kv_k-y_n|<1/m$.
    $endgroup$
    – Anthony Quas
    Apr 21 at 19:18


















$begingroup$
and if they are iid the probability of being in a closed hyperplane $(h)^perp$ is $P(v_kperp h, k=1,2,dots)=0$
$endgroup$
– Pietro Majer
Apr 21 at 17:07




$begingroup$
and if they are iid the probability of being in a closed hyperplane $(h)^perp$ is $P(v_kperp h, k=1,2,dots)=0$
$endgroup$
– Pietro Majer
Apr 21 at 17:07




3




3




$begingroup$
@Pietro Majer: this is the probability that the vectors all lie in a given closed hyperplane.
$endgroup$
– Anthony Quas
Apr 21 at 17:10






$begingroup$
@Pietro Majer: this is the probability that the vectors all lie in a given closed hyperplane.
$endgroup$
– Anthony Quas
Apr 21 at 17:10






3




3




$begingroup$
Whoever voted to close this, I'm pretty sure you don't understand the question. This is subtle and interesting.
$endgroup$
– Anthony Quas
Apr 21 at 17:18




$begingroup$
Whoever voted to close this, I'm pretty sure you don't understand the question. This is subtle and interesting.
$endgroup$
– Anthony Quas
Apr 21 at 17:18




1




1




$begingroup$
Maybe I'm missing something obvious, but is it even clear that the event you are interested in is measurable?
$endgroup$
– Jochen Glueck
Apr 21 at 17:21






$begingroup$
Maybe I'm missing something obvious, but is it even clear that the event you are interested in is measurable?
$endgroup$
– Jochen Glueck
Apr 21 at 17:21






1




1




$begingroup$
@JochenGlueck: Yes it's measurable: Let $(y_n)$ be a dense sequence in $mathcal H$. Then the event is: for all $m>0$, for all $n>0$, there exist $k>0$ and rational $t_1,ldots,t_k$ such that $|t_1v_1+ldots+t_kv_k-y_n|<1/m$.
$endgroup$
– Anthony Quas
Apr 21 at 19:18






$begingroup$
@JochenGlueck: Yes it's measurable: Let $(y_n)$ be a dense sequence in $mathcal H$. Then the event is: for all $m>0$, for all $n>0$, there exist $k>0$ and rational $t_1,ldots,t_k$ such that $|t_1v_1+ldots+t_kv_k-y_n|<1/m$.
$endgroup$
– Anthony Quas
Apr 21 at 19:18












2 Answers
2






active

oldest

votes


















6












$begingroup$

(This may turn out to be a simplified version of J. E. Pascoe's answer).



The support of (the distribution of) $v$, that we denote by $operatorname{supp} v$, is the set of vectors $h in mathcal{H}$ such that $P(v in B(h, varepsilon)) > 0$ for every $varepsilon > 0$. We list some properties of this set.




  1. The set $operatorname{supp} v$ is the complement of the union of all open sets $B$ such that $P(v in B) = 0$. Thus, the support is a closed set.


  2. Since $mathcal{H}$ is a separable metric space, it has a countable topological base $mathcal{B}$, and $operatorname{supp} v$ is the complement of the union of all $B in mathcal{B}$ such that $P(v in B) = 0$. By countable additivity, it follows that $P(v in operatorname{supp} v) = 1$ (the support is a set of full measure).


  3. With probability one, the closure of the random set $V = {v_1, v_2, ldots}$ contains $operatorname{supp} v$. Indeed, let ${h_1, h_2, ldots}$ be a countable, dense subset of $operatorname{supp} v$. For every $i, n = 1, 2, ldots$ we have $P(v in B(h_i, tfrac{1}{n})) > 0$, and thus, by Borel–Cantelli, $P(V cap B(h_i, tfrac{1}{n}) = varnothing) = 0$. It follows that $h_i in overline{V}$ for every $i = 1, 2, ldots$, and consequently $operatorname{supp} v subseteq overline{V}$.


  4. For every $h in mathcal{H}$, we have $P(h perp v) < 1$, and therefore $h$ is not orthogonal to $operatorname{supp} v$. It follows that the closed span of $operatorname{supp} v$ is $mathcal{H}$.



It remains to note that the closed span of $V$ is the same as the closed span of the closure of $V$, which with probability one contains the closed span of $operatorname{supp} v$, which we have shown to be equal to $mathcal{H}$.



(Item 1 is valid for any topological space; items 2 and 3 work in an arbitrary separable metric space.)






share|cite|improve this answer











$endgroup$













  • $begingroup$
    It would perhaps be even clearer to say that for each fixed vector $hintextrm{supp}; v$, we have $P(hin overline{V})=1$. (This shows that the assumption that $H$ is separable is used, and how you avoid the pitfall from Pietro's comment above.)
    $endgroup$
    – Christian Remling
    Apr 22 at 19:53












  • $begingroup$
    @ChristianRemling: I edited my answer to emphasize where separability is essential. Let me know if anything is wrong. Thanks!
    $endgroup$
    – Mateusz Kwaśnicki
    2 days ago










  • $begingroup$
    Looks good to me. I didn't mean to suggest that such an extensive edit was needed, of course, just wanted to draw attention to where your answer differs from Pietro's comment.
    $endgroup$
    – Christian Remling
    2 days ago



















0












$begingroup$

Another Try



We say a $mathcal{H}$-valued random variable $h$ is a random vector if $P(h perp g)<1$ for all $gin mathcal{H}.$



If $h_1, h_2, ldots$ is a sequence independent identically distributed of random vectors,
then, almost surely, the closed span of the $h_i$ is equal to $mathcal{H}.$



First we will need a lemma.



Lemma 1
Let $h$ be a random vector.
There is a countable subset $A$ of $mathcal{H}$ such that the closed span of the elements of $A$ is equal to $mathcal{H}$
and for every point $ain A,$ $P(hin U)>0$ for any neighborhood $U$ of $a.$



Proof
For any subset $A$ such that for every point $ain A,$ $P(hin U)>0$ for any neighborhood $U$ of $a,$ and
the closed span of the elements of $A$ is not equal to $mathcal{H},$
we will show that we can grow $A$ by a single element which is not in closed span of the elements of $A.$
We can only do this a countable number of times because the Hilbert space dimension of $mathcal{H}$ is countable.
(Otherwise, via Gram-Schmidt, we could construct an uncountable orthonormal set by transfinite induction.)



Choose $g$ such that $g perp a$ for all $ain A.$ Now, $P(h perp g)<1.$ So there must be a point $b$ such that
$P(hin U) >0$ for every neighborhood of $b$ and $b$ is not perpendicular to $g,$ therefore, $b$ is not in the span of the elements of $A.$ QED



Suppose $h_1, h_2, ldots$ is a sequence independent identically distributed of random vectors.
Let $A$ be as in Lemma 1. Index $A$ a a sequence $a_n.$
Let $B_{m,n}$ be a ball of radius $1/m$ centered at $a_n$
Almost surely, the sequence $h_i$ must visit $B_{m,n}$ infinitely often,
as $P(h_iin B_{m,n})>0$. Therefore $A$ is a subset of the closure of the values of the sequence. (We have essentially the fact that a random function $f:mathbb{N}rightarrow mathbb{N}^2$ is surjective with infinite multiplicity.)






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$endgroup$













  • $begingroup$
    The point is that you "keep going" by transfinite induction. The process must stop at some countable ordinal before $omega_1$ as the space is countable dimensional.
    $endgroup$
    – J. E. Pascoe
    Apr 21 at 19:59










  • $begingroup$
    That is for each $alpha < omega_1$ there would be $A_alpha$ such that if $alpha < beta,$ $A_alpha^perp cap A_beta neq {0}.$
    $endgroup$
    – J. E. Pascoe
    Apr 21 at 20:06












  • $begingroup$
    "Let $A$ be as in Lemma 1" ... "we may take $A$ to consist of only isolated points, as in a Polish space a countable closed set is equal to the closure of its isolated points." But Lemma 1 does not guarantee that $A$ is closed.
    $endgroup$
    – Iosif Pinelis
    Apr 21 at 20:07












  • $begingroup$
    That does seem to be a gap @IosifPinelis . Ideas for closing it?
    $endgroup$
    – J. E. Pascoe
    Apr 21 at 20:10












  • $begingroup$
    Could you elaborate a bit further on why the set $A$ in Lemma 1 is countable? Of course, there exist linearly independent sets in $H$ that are uncountable -- just choose your favourite Hamel basis of $H$.
    $endgroup$
    – Jochen Glueck
    Apr 21 at 20:18














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2 Answers
2






active

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2 Answers
2






active

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votes









active

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votes






active

oldest

votes









6












$begingroup$

(This may turn out to be a simplified version of J. E. Pascoe's answer).



The support of (the distribution of) $v$, that we denote by $operatorname{supp} v$, is the set of vectors $h in mathcal{H}$ such that $P(v in B(h, varepsilon)) > 0$ for every $varepsilon > 0$. We list some properties of this set.




  1. The set $operatorname{supp} v$ is the complement of the union of all open sets $B$ such that $P(v in B) = 0$. Thus, the support is a closed set.


  2. Since $mathcal{H}$ is a separable metric space, it has a countable topological base $mathcal{B}$, and $operatorname{supp} v$ is the complement of the union of all $B in mathcal{B}$ such that $P(v in B) = 0$. By countable additivity, it follows that $P(v in operatorname{supp} v) = 1$ (the support is a set of full measure).


  3. With probability one, the closure of the random set $V = {v_1, v_2, ldots}$ contains $operatorname{supp} v$. Indeed, let ${h_1, h_2, ldots}$ be a countable, dense subset of $operatorname{supp} v$. For every $i, n = 1, 2, ldots$ we have $P(v in B(h_i, tfrac{1}{n})) > 0$, and thus, by Borel–Cantelli, $P(V cap B(h_i, tfrac{1}{n}) = varnothing) = 0$. It follows that $h_i in overline{V}$ for every $i = 1, 2, ldots$, and consequently $operatorname{supp} v subseteq overline{V}$.


  4. For every $h in mathcal{H}$, we have $P(h perp v) < 1$, and therefore $h$ is not orthogonal to $operatorname{supp} v$. It follows that the closed span of $operatorname{supp} v$ is $mathcal{H}$.



It remains to note that the closed span of $V$ is the same as the closed span of the closure of $V$, which with probability one contains the closed span of $operatorname{supp} v$, which we have shown to be equal to $mathcal{H}$.



(Item 1 is valid for any topological space; items 2 and 3 work in an arbitrary separable metric space.)






share|cite|improve this answer











$endgroup$













  • $begingroup$
    It would perhaps be even clearer to say that for each fixed vector $hintextrm{supp}; v$, we have $P(hin overline{V})=1$. (This shows that the assumption that $H$ is separable is used, and how you avoid the pitfall from Pietro's comment above.)
    $endgroup$
    – Christian Remling
    Apr 22 at 19:53












  • $begingroup$
    @ChristianRemling: I edited my answer to emphasize where separability is essential. Let me know if anything is wrong. Thanks!
    $endgroup$
    – Mateusz Kwaśnicki
    2 days ago










  • $begingroup$
    Looks good to me. I didn't mean to suggest that such an extensive edit was needed, of course, just wanted to draw attention to where your answer differs from Pietro's comment.
    $endgroup$
    – Christian Remling
    2 days ago
















6












$begingroup$

(This may turn out to be a simplified version of J. E. Pascoe's answer).



The support of (the distribution of) $v$, that we denote by $operatorname{supp} v$, is the set of vectors $h in mathcal{H}$ such that $P(v in B(h, varepsilon)) > 0$ for every $varepsilon > 0$. We list some properties of this set.




  1. The set $operatorname{supp} v$ is the complement of the union of all open sets $B$ such that $P(v in B) = 0$. Thus, the support is a closed set.


  2. Since $mathcal{H}$ is a separable metric space, it has a countable topological base $mathcal{B}$, and $operatorname{supp} v$ is the complement of the union of all $B in mathcal{B}$ such that $P(v in B) = 0$. By countable additivity, it follows that $P(v in operatorname{supp} v) = 1$ (the support is a set of full measure).


  3. With probability one, the closure of the random set $V = {v_1, v_2, ldots}$ contains $operatorname{supp} v$. Indeed, let ${h_1, h_2, ldots}$ be a countable, dense subset of $operatorname{supp} v$. For every $i, n = 1, 2, ldots$ we have $P(v in B(h_i, tfrac{1}{n})) > 0$, and thus, by Borel–Cantelli, $P(V cap B(h_i, tfrac{1}{n}) = varnothing) = 0$. It follows that $h_i in overline{V}$ for every $i = 1, 2, ldots$, and consequently $operatorname{supp} v subseteq overline{V}$.


  4. For every $h in mathcal{H}$, we have $P(h perp v) < 1$, and therefore $h$ is not orthogonal to $operatorname{supp} v$. It follows that the closed span of $operatorname{supp} v$ is $mathcal{H}$.



It remains to note that the closed span of $V$ is the same as the closed span of the closure of $V$, which with probability one contains the closed span of $operatorname{supp} v$, which we have shown to be equal to $mathcal{H}$.



(Item 1 is valid for any topological space; items 2 and 3 work in an arbitrary separable metric space.)






share|cite|improve this answer











$endgroup$













  • $begingroup$
    It would perhaps be even clearer to say that for each fixed vector $hintextrm{supp}; v$, we have $P(hin overline{V})=1$. (This shows that the assumption that $H$ is separable is used, and how you avoid the pitfall from Pietro's comment above.)
    $endgroup$
    – Christian Remling
    Apr 22 at 19:53












  • $begingroup$
    @ChristianRemling: I edited my answer to emphasize where separability is essential. Let me know if anything is wrong. Thanks!
    $endgroup$
    – Mateusz Kwaśnicki
    2 days ago










  • $begingroup$
    Looks good to me. I didn't mean to suggest that such an extensive edit was needed, of course, just wanted to draw attention to where your answer differs from Pietro's comment.
    $endgroup$
    – Christian Remling
    2 days ago














6












6








6





$begingroup$

(This may turn out to be a simplified version of J. E. Pascoe's answer).



The support of (the distribution of) $v$, that we denote by $operatorname{supp} v$, is the set of vectors $h in mathcal{H}$ such that $P(v in B(h, varepsilon)) > 0$ for every $varepsilon > 0$. We list some properties of this set.




  1. The set $operatorname{supp} v$ is the complement of the union of all open sets $B$ such that $P(v in B) = 0$. Thus, the support is a closed set.


  2. Since $mathcal{H}$ is a separable metric space, it has a countable topological base $mathcal{B}$, and $operatorname{supp} v$ is the complement of the union of all $B in mathcal{B}$ such that $P(v in B) = 0$. By countable additivity, it follows that $P(v in operatorname{supp} v) = 1$ (the support is a set of full measure).


  3. With probability one, the closure of the random set $V = {v_1, v_2, ldots}$ contains $operatorname{supp} v$. Indeed, let ${h_1, h_2, ldots}$ be a countable, dense subset of $operatorname{supp} v$. For every $i, n = 1, 2, ldots$ we have $P(v in B(h_i, tfrac{1}{n})) > 0$, and thus, by Borel–Cantelli, $P(V cap B(h_i, tfrac{1}{n}) = varnothing) = 0$. It follows that $h_i in overline{V}$ for every $i = 1, 2, ldots$, and consequently $operatorname{supp} v subseteq overline{V}$.


  4. For every $h in mathcal{H}$, we have $P(h perp v) < 1$, and therefore $h$ is not orthogonal to $operatorname{supp} v$. It follows that the closed span of $operatorname{supp} v$ is $mathcal{H}$.



It remains to note that the closed span of $V$ is the same as the closed span of the closure of $V$, which with probability one contains the closed span of $operatorname{supp} v$, which we have shown to be equal to $mathcal{H}$.



(Item 1 is valid for any topological space; items 2 and 3 work in an arbitrary separable metric space.)






share|cite|improve this answer











$endgroup$



(This may turn out to be a simplified version of J. E. Pascoe's answer).



The support of (the distribution of) $v$, that we denote by $operatorname{supp} v$, is the set of vectors $h in mathcal{H}$ such that $P(v in B(h, varepsilon)) > 0$ for every $varepsilon > 0$. We list some properties of this set.




  1. The set $operatorname{supp} v$ is the complement of the union of all open sets $B$ such that $P(v in B) = 0$. Thus, the support is a closed set.


  2. Since $mathcal{H}$ is a separable metric space, it has a countable topological base $mathcal{B}$, and $operatorname{supp} v$ is the complement of the union of all $B in mathcal{B}$ such that $P(v in B) = 0$. By countable additivity, it follows that $P(v in operatorname{supp} v) = 1$ (the support is a set of full measure).


  3. With probability one, the closure of the random set $V = {v_1, v_2, ldots}$ contains $operatorname{supp} v$. Indeed, let ${h_1, h_2, ldots}$ be a countable, dense subset of $operatorname{supp} v$. For every $i, n = 1, 2, ldots$ we have $P(v in B(h_i, tfrac{1}{n})) > 0$, and thus, by Borel–Cantelli, $P(V cap B(h_i, tfrac{1}{n}) = varnothing) = 0$. It follows that $h_i in overline{V}$ for every $i = 1, 2, ldots$, and consequently $operatorname{supp} v subseteq overline{V}$.


  4. For every $h in mathcal{H}$, we have $P(h perp v) < 1$, and therefore $h$ is not orthogonal to $operatorname{supp} v$. It follows that the closed span of $operatorname{supp} v$ is $mathcal{H}$.



It remains to note that the closed span of $V$ is the same as the closed span of the closure of $V$, which with probability one contains the closed span of $operatorname{supp} v$, which we have shown to be equal to $mathcal{H}$.



(Item 1 is valid for any topological space; items 2 and 3 work in an arbitrary separable metric space.)







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 2 days ago

























answered Apr 21 at 21:40









Mateusz KwaśnickiMateusz Kwaśnicki

4,7421619




4,7421619












  • $begingroup$
    It would perhaps be even clearer to say that for each fixed vector $hintextrm{supp}; v$, we have $P(hin overline{V})=1$. (This shows that the assumption that $H$ is separable is used, and how you avoid the pitfall from Pietro's comment above.)
    $endgroup$
    – Christian Remling
    Apr 22 at 19:53












  • $begingroup$
    @ChristianRemling: I edited my answer to emphasize where separability is essential. Let me know if anything is wrong. Thanks!
    $endgroup$
    – Mateusz Kwaśnicki
    2 days ago










  • $begingroup$
    Looks good to me. I didn't mean to suggest that such an extensive edit was needed, of course, just wanted to draw attention to where your answer differs from Pietro's comment.
    $endgroup$
    – Christian Remling
    2 days ago


















  • $begingroup$
    It would perhaps be even clearer to say that for each fixed vector $hintextrm{supp}; v$, we have $P(hin overline{V})=1$. (This shows that the assumption that $H$ is separable is used, and how you avoid the pitfall from Pietro's comment above.)
    $endgroup$
    – Christian Remling
    Apr 22 at 19:53












  • $begingroup$
    @ChristianRemling: I edited my answer to emphasize where separability is essential. Let me know if anything is wrong. Thanks!
    $endgroup$
    – Mateusz Kwaśnicki
    2 days ago










  • $begingroup$
    Looks good to me. I didn't mean to suggest that such an extensive edit was needed, of course, just wanted to draw attention to where your answer differs from Pietro's comment.
    $endgroup$
    – Christian Remling
    2 days ago
















$begingroup$
It would perhaps be even clearer to say that for each fixed vector $hintextrm{supp}; v$, we have $P(hin overline{V})=1$. (This shows that the assumption that $H$ is separable is used, and how you avoid the pitfall from Pietro's comment above.)
$endgroup$
– Christian Remling
Apr 22 at 19:53






$begingroup$
It would perhaps be even clearer to say that for each fixed vector $hintextrm{supp}; v$, we have $P(hin overline{V})=1$. (This shows that the assumption that $H$ is separable is used, and how you avoid the pitfall from Pietro's comment above.)
$endgroup$
– Christian Remling
Apr 22 at 19:53














$begingroup$
@ChristianRemling: I edited my answer to emphasize where separability is essential. Let me know if anything is wrong. Thanks!
$endgroup$
– Mateusz Kwaśnicki
2 days ago




$begingroup$
@ChristianRemling: I edited my answer to emphasize where separability is essential. Let me know if anything is wrong. Thanks!
$endgroup$
– Mateusz Kwaśnicki
2 days ago












$begingroup$
Looks good to me. I didn't mean to suggest that such an extensive edit was needed, of course, just wanted to draw attention to where your answer differs from Pietro's comment.
$endgroup$
– Christian Remling
2 days ago




$begingroup$
Looks good to me. I didn't mean to suggest that such an extensive edit was needed, of course, just wanted to draw attention to where your answer differs from Pietro's comment.
$endgroup$
– Christian Remling
2 days ago











0












$begingroup$

Another Try



We say a $mathcal{H}$-valued random variable $h$ is a random vector if $P(h perp g)<1$ for all $gin mathcal{H}.$



If $h_1, h_2, ldots$ is a sequence independent identically distributed of random vectors,
then, almost surely, the closed span of the $h_i$ is equal to $mathcal{H}.$



First we will need a lemma.



Lemma 1
Let $h$ be a random vector.
There is a countable subset $A$ of $mathcal{H}$ such that the closed span of the elements of $A$ is equal to $mathcal{H}$
and for every point $ain A,$ $P(hin U)>0$ for any neighborhood $U$ of $a.$



Proof
For any subset $A$ such that for every point $ain A,$ $P(hin U)>0$ for any neighborhood $U$ of $a,$ and
the closed span of the elements of $A$ is not equal to $mathcal{H},$
we will show that we can grow $A$ by a single element which is not in closed span of the elements of $A.$
We can only do this a countable number of times because the Hilbert space dimension of $mathcal{H}$ is countable.
(Otherwise, via Gram-Schmidt, we could construct an uncountable orthonormal set by transfinite induction.)



Choose $g$ such that $g perp a$ for all $ain A.$ Now, $P(h perp g)<1.$ So there must be a point $b$ such that
$P(hin U) >0$ for every neighborhood of $b$ and $b$ is not perpendicular to $g,$ therefore, $b$ is not in the span of the elements of $A.$ QED



Suppose $h_1, h_2, ldots$ is a sequence independent identically distributed of random vectors.
Let $A$ be as in Lemma 1. Index $A$ a a sequence $a_n.$
Let $B_{m,n}$ be a ball of radius $1/m$ centered at $a_n$
Almost surely, the sequence $h_i$ must visit $B_{m,n}$ infinitely often,
as $P(h_iin B_{m,n})>0$. Therefore $A$ is a subset of the closure of the values of the sequence. (We have essentially the fact that a random function $f:mathbb{N}rightarrow mathbb{N}^2$ is surjective with infinite multiplicity.)






share|cite|improve this answer











$endgroup$













  • $begingroup$
    The point is that you "keep going" by transfinite induction. The process must stop at some countable ordinal before $omega_1$ as the space is countable dimensional.
    $endgroup$
    – J. E. Pascoe
    Apr 21 at 19:59










  • $begingroup$
    That is for each $alpha < omega_1$ there would be $A_alpha$ such that if $alpha < beta,$ $A_alpha^perp cap A_beta neq {0}.$
    $endgroup$
    – J. E. Pascoe
    Apr 21 at 20:06












  • $begingroup$
    "Let $A$ be as in Lemma 1" ... "we may take $A$ to consist of only isolated points, as in a Polish space a countable closed set is equal to the closure of its isolated points." But Lemma 1 does not guarantee that $A$ is closed.
    $endgroup$
    – Iosif Pinelis
    Apr 21 at 20:07












  • $begingroup$
    That does seem to be a gap @IosifPinelis . Ideas for closing it?
    $endgroup$
    – J. E. Pascoe
    Apr 21 at 20:10












  • $begingroup$
    Could you elaborate a bit further on why the set $A$ in Lemma 1 is countable? Of course, there exist linearly independent sets in $H$ that are uncountable -- just choose your favourite Hamel basis of $H$.
    $endgroup$
    – Jochen Glueck
    Apr 21 at 20:18


















0












$begingroup$

Another Try



We say a $mathcal{H}$-valued random variable $h$ is a random vector if $P(h perp g)<1$ for all $gin mathcal{H}.$



If $h_1, h_2, ldots$ is a sequence independent identically distributed of random vectors,
then, almost surely, the closed span of the $h_i$ is equal to $mathcal{H}.$



First we will need a lemma.



Lemma 1
Let $h$ be a random vector.
There is a countable subset $A$ of $mathcal{H}$ such that the closed span of the elements of $A$ is equal to $mathcal{H}$
and for every point $ain A,$ $P(hin U)>0$ for any neighborhood $U$ of $a.$



Proof
For any subset $A$ such that for every point $ain A,$ $P(hin U)>0$ for any neighborhood $U$ of $a,$ and
the closed span of the elements of $A$ is not equal to $mathcal{H},$
we will show that we can grow $A$ by a single element which is not in closed span of the elements of $A.$
We can only do this a countable number of times because the Hilbert space dimension of $mathcal{H}$ is countable.
(Otherwise, via Gram-Schmidt, we could construct an uncountable orthonormal set by transfinite induction.)



Choose $g$ such that $g perp a$ for all $ain A.$ Now, $P(h perp g)<1.$ So there must be a point $b$ such that
$P(hin U) >0$ for every neighborhood of $b$ and $b$ is not perpendicular to $g,$ therefore, $b$ is not in the span of the elements of $A.$ QED



Suppose $h_1, h_2, ldots$ is a sequence independent identically distributed of random vectors.
Let $A$ be as in Lemma 1. Index $A$ a a sequence $a_n.$
Let $B_{m,n}$ be a ball of radius $1/m$ centered at $a_n$
Almost surely, the sequence $h_i$ must visit $B_{m,n}$ infinitely often,
as $P(h_iin B_{m,n})>0$. Therefore $A$ is a subset of the closure of the values of the sequence. (We have essentially the fact that a random function $f:mathbb{N}rightarrow mathbb{N}^2$ is surjective with infinite multiplicity.)






share|cite|improve this answer











$endgroup$













  • $begingroup$
    The point is that you "keep going" by transfinite induction. The process must stop at some countable ordinal before $omega_1$ as the space is countable dimensional.
    $endgroup$
    – J. E. Pascoe
    Apr 21 at 19:59










  • $begingroup$
    That is for each $alpha < omega_1$ there would be $A_alpha$ such that if $alpha < beta,$ $A_alpha^perp cap A_beta neq {0}.$
    $endgroup$
    – J. E. Pascoe
    Apr 21 at 20:06












  • $begingroup$
    "Let $A$ be as in Lemma 1" ... "we may take $A$ to consist of only isolated points, as in a Polish space a countable closed set is equal to the closure of its isolated points." But Lemma 1 does not guarantee that $A$ is closed.
    $endgroup$
    – Iosif Pinelis
    Apr 21 at 20:07












  • $begingroup$
    That does seem to be a gap @IosifPinelis . Ideas for closing it?
    $endgroup$
    – J. E. Pascoe
    Apr 21 at 20:10












  • $begingroup$
    Could you elaborate a bit further on why the set $A$ in Lemma 1 is countable? Of course, there exist linearly independent sets in $H$ that are uncountable -- just choose your favourite Hamel basis of $H$.
    $endgroup$
    – Jochen Glueck
    Apr 21 at 20:18
















0












0








0





$begingroup$

Another Try



We say a $mathcal{H}$-valued random variable $h$ is a random vector if $P(h perp g)<1$ for all $gin mathcal{H}.$



If $h_1, h_2, ldots$ is a sequence independent identically distributed of random vectors,
then, almost surely, the closed span of the $h_i$ is equal to $mathcal{H}.$



First we will need a lemma.



Lemma 1
Let $h$ be a random vector.
There is a countable subset $A$ of $mathcal{H}$ such that the closed span of the elements of $A$ is equal to $mathcal{H}$
and for every point $ain A,$ $P(hin U)>0$ for any neighborhood $U$ of $a.$



Proof
For any subset $A$ such that for every point $ain A,$ $P(hin U)>0$ for any neighborhood $U$ of $a,$ and
the closed span of the elements of $A$ is not equal to $mathcal{H},$
we will show that we can grow $A$ by a single element which is not in closed span of the elements of $A.$
We can only do this a countable number of times because the Hilbert space dimension of $mathcal{H}$ is countable.
(Otherwise, via Gram-Schmidt, we could construct an uncountable orthonormal set by transfinite induction.)



Choose $g$ such that $g perp a$ for all $ain A.$ Now, $P(h perp g)<1.$ So there must be a point $b$ such that
$P(hin U) >0$ for every neighborhood of $b$ and $b$ is not perpendicular to $g,$ therefore, $b$ is not in the span of the elements of $A.$ QED



Suppose $h_1, h_2, ldots$ is a sequence independent identically distributed of random vectors.
Let $A$ be as in Lemma 1. Index $A$ a a sequence $a_n.$
Let $B_{m,n}$ be a ball of radius $1/m$ centered at $a_n$
Almost surely, the sequence $h_i$ must visit $B_{m,n}$ infinitely often,
as $P(h_iin B_{m,n})>0$. Therefore $A$ is a subset of the closure of the values of the sequence. (We have essentially the fact that a random function $f:mathbb{N}rightarrow mathbb{N}^2$ is surjective with infinite multiplicity.)






share|cite|improve this answer











$endgroup$



Another Try



We say a $mathcal{H}$-valued random variable $h$ is a random vector if $P(h perp g)<1$ for all $gin mathcal{H}.$



If $h_1, h_2, ldots$ is a sequence independent identically distributed of random vectors,
then, almost surely, the closed span of the $h_i$ is equal to $mathcal{H}.$



First we will need a lemma.



Lemma 1
Let $h$ be a random vector.
There is a countable subset $A$ of $mathcal{H}$ such that the closed span of the elements of $A$ is equal to $mathcal{H}$
and for every point $ain A,$ $P(hin U)>0$ for any neighborhood $U$ of $a.$



Proof
For any subset $A$ such that for every point $ain A,$ $P(hin U)>0$ for any neighborhood $U$ of $a,$ and
the closed span of the elements of $A$ is not equal to $mathcal{H},$
we will show that we can grow $A$ by a single element which is not in closed span of the elements of $A.$
We can only do this a countable number of times because the Hilbert space dimension of $mathcal{H}$ is countable.
(Otherwise, via Gram-Schmidt, we could construct an uncountable orthonormal set by transfinite induction.)



Choose $g$ such that $g perp a$ for all $ain A.$ Now, $P(h perp g)<1.$ So there must be a point $b$ such that
$P(hin U) >0$ for every neighborhood of $b$ and $b$ is not perpendicular to $g,$ therefore, $b$ is not in the span of the elements of $A.$ QED



Suppose $h_1, h_2, ldots$ is a sequence independent identically distributed of random vectors.
Let $A$ be as in Lemma 1. Index $A$ a a sequence $a_n.$
Let $B_{m,n}$ be a ball of radius $1/m$ centered at $a_n$
Almost surely, the sequence $h_i$ must visit $B_{m,n}$ infinitely often,
as $P(h_iin B_{m,n})>0$. Therefore $A$ is a subset of the closure of the values of the sequence. (We have essentially the fact that a random function $f:mathbb{N}rightarrow mathbb{N}^2$ is surjective with infinite multiplicity.)







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Apr 21 at 21:24

























answered Apr 21 at 19:48









J. E. PascoeJ. E. Pascoe

575316




575316












  • $begingroup$
    The point is that you "keep going" by transfinite induction. The process must stop at some countable ordinal before $omega_1$ as the space is countable dimensional.
    $endgroup$
    – J. E. Pascoe
    Apr 21 at 19:59










  • $begingroup$
    That is for each $alpha < omega_1$ there would be $A_alpha$ such that if $alpha < beta,$ $A_alpha^perp cap A_beta neq {0}.$
    $endgroup$
    – J. E. Pascoe
    Apr 21 at 20:06












  • $begingroup$
    "Let $A$ be as in Lemma 1" ... "we may take $A$ to consist of only isolated points, as in a Polish space a countable closed set is equal to the closure of its isolated points." But Lemma 1 does not guarantee that $A$ is closed.
    $endgroup$
    – Iosif Pinelis
    Apr 21 at 20:07












  • $begingroup$
    That does seem to be a gap @IosifPinelis . Ideas for closing it?
    $endgroup$
    – J. E. Pascoe
    Apr 21 at 20:10












  • $begingroup$
    Could you elaborate a bit further on why the set $A$ in Lemma 1 is countable? Of course, there exist linearly independent sets in $H$ that are uncountable -- just choose your favourite Hamel basis of $H$.
    $endgroup$
    – Jochen Glueck
    Apr 21 at 20:18




















  • $begingroup$
    The point is that you "keep going" by transfinite induction. The process must stop at some countable ordinal before $omega_1$ as the space is countable dimensional.
    $endgroup$
    – J. E. Pascoe
    Apr 21 at 19:59










  • $begingroup$
    That is for each $alpha < omega_1$ there would be $A_alpha$ such that if $alpha < beta,$ $A_alpha^perp cap A_beta neq {0}.$
    $endgroup$
    – J. E. Pascoe
    Apr 21 at 20:06












  • $begingroup$
    "Let $A$ be as in Lemma 1" ... "we may take $A$ to consist of only isolated points, as in a Polish space a countable closed set is equal to the closure of its isolated points." But Lemma 1 does not guarantee that $A$ is closed.
    $endgroup$
    – Iosif Pinelis
    Apr 21 at 20:07












  • $begingroup$
    That does seem to be a gap @IosifPinelis . Ideas for closing it?
    $endgroup$
    – J. E. Pascoe
    Apr 21 at 20:10












  • $begingroup$
    Could you elaborate a bit further on why the set $A$ in Lemma 1 is countable? Of course, there exist linearly independent sets in $H$ that are uncountable -- just choose your favourite Hamel basis of $H$.
    $endgroup$
    – Jochen Glueck
    Apr 21 at 20:18


















$begingroup$
The point is that you "keep going" by transfinite induction. The process must stop at some countable ordinal before $omega_1$ as the space is countable dimensional.
$endgroup$
– J. E. Pascoe
Apr 21 at 19:59




$begingroup$
The point is that you "keep going" by transfinite induction. The process must stop at some countable ordinal before $omega_1$ as the space is countable dimensional.
$endgroup$
– J. E. Pascoe
Apr 21 at 19:59












$begingroup$
That is for each $alpha < omega_1$ there would be $A_alpha$ such that if $alpha < beta,$ $A_alpha^perp cap A_beta neq {0}.$
$endgroup$
– J. E. Pascoe
Apr 21 at 20:06






$begingroup$
That is for each $alpha < omega_1$ there would be $A_alpha$ such that if $alpha < beta,$ $A_alpha^perp cap A_beta neq {0}.$
$endgroup$
– J. E. Pascoe
Apr 21 at 20:06














$begingroup$
"Let $A$ be as in Lemma 1" ... "we may take $A$ to consist of only isolated points, as in a Polish space a countable closed set is equal to the closure of its isolated points." But Lemma 1 does not guarantee that $A$ is closed.
$endgroup$
– Iosif Pinelis
Apr 21 at 20:07






$begingroup$
"Let $A$ be as in Lemma 1" ... "we may take $A$ to consist of only isolated points, as in a Polish space a countable closed set is equal to the closure of its isolated points." But Lemma 1 does not guarantee that $A$ is closed.
$endgroup$
– Iosif Pinelis
Apr 21 at 20:07














$begingroup$
That does seem to be a gap @IosifPinelis . Ideas for closing it?
$endgroup$
– J. E. Pascoe
Apr 21 at 20:10






$begingroup$
That does seem to be a gap @IosifPinelis . Ideas for closing it?
$endgroup$
– J. E. Pascoe
Apr 21 at 20:10














$begingroup$
Could you elaborate a bit further on why the set $A$ in Lemma 1 is countable? Of course, there exist linearly independent sets in $H$ that are uncountable -- just choose your favourite Hamel basis of $H$.
$endgroup$
– Jochen Glueck
Apr 21 at 20:18






$begingroup$
Could you elaborate a bit further on why the set $A$ in Lemma 1 is countable? Of course, there exist linearly independent sets in $H$ that are uncountable -- just choose your favourite Hamel basis of $H$.
$endgroup$
– Jochen Glueck
Apr 21 at 20:18




















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