What is the topology on the direct limit of $Bbb R^i rightarrow Bbb R^j$












1












$begingroup$


If we let $X$ be direct limit of
$$Bbb R rightarrow Bbb R^2 rightarrow Bbb R^3 rightarrow cdots $$
where each arrow is inclusion to first coordinates.



What is a choice of $X$ (with a toplogy that coincides with the direct limit topology).



EDIT: I believe it is subspace of $Bbb R^{Bbb N}$, sequence eventually $0$ taking values in $Bbb R$, with the product topology. I wonder if this is correct.






general-topology algebraic-topology







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$endgroup$








  • 2




    $begingroup$
    The direct limit is the set of sequences which are eventually zero.
    $endgroup$
    – Lord Shark the Unknown
    Dec 4 '18 at 7:40










  • $begingroup$
    Yes, sorry, that is right. But how about the topology?
    $endgroup$
    – CL.
    Dec 4 '18 at 7:41


















1












$begingroup$


If we let $X$ be direct limit of
$$Bbb R rightarrow Bbb R^2 rightarrow Bbb R^3 rightarrow cdots $$
where each arrow is inclusion to first coordinates.



What is a choice of $X$ (with a toplogy that coincides with the direct limit topology).



EDIT: I believe it is subspace of $Bbb R^{Bbb N}$, sequence eventually $0$ taking values in $Bbb R$, with the product topology. I wonder if this is correct.






general-topology algebraic-topology







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    The direct limit is the set of sequences which are eventually zero.
    $endgroup$
    – Lord Shark the Unknown
    Dec 4 '18 at 7:40










  • $begingroup$
    Yes, sorry, that is right. But how about the topology?
    $endgroup$
    – CL.
    Dec 4 '18 at 7:41
















1












1








1


1



$begingroup$


If we let $X$ be direct limit of
$$Bbb R rightarrow Bbb R^2 rightarrow Bbb R^3 rightarrow cdots $$
where each arrow is inclusion to first coordinates.



What is a choice of $X$ (with a toplogy that coincides with the direct limit topology).



EDIT: I believe it is subspace of $Bbb R^{Bbb N}$, sequence eventually $0$ taking values in $Bbb R$, with the product topology. I wonder if this is correct.






general-topology algebraic-topology







share|cite|improve this question











$endgroup$




If we let $X$ be direct limit of
$$Bbb R rightarrow Bbb R^2 rightarrow Bbb R^3 rightarrow cdots $$
where each arrow is inclusion to first coordinates.



What is a choice of $X$ (with a toplogy that coincides with the direct limit topology).



EDIT: I believe it is subspace of $Bbb R^{Bbb N}$, sequence eventually $0$ taking values in $Bbb R$, with the product topology. I wonder if this is correct.






general-topology algebraic-topology




general-topology algebraic-topology






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 6 '18 at 9:18







CL.

















asked Dec 4 '18 at 7:39









CL.CL.

2,2232825




2,2232825








  • 2




    $begingroup$
    The direct limit is the set of sequences which are eventually zero.
    $endgroup$
    – Lord Shark the Unknown
    Dec 4 '18 at 7:40










  • $begingroup$
    Yes, sorry, that is right. But how about the topology?
    $endgroup$
    – CL.
    Dec 4 '18 at 7:41
















  • 2




    $begingroup$
    The direct limit is the set of sequences which are eventually zero.
    $endgroup$
    – Lord Shark the Unknown
    Dec 4 '18 at 7:40










  • $begingroup$
    Yes, sorry, that is right. But how about the topology?
    $endgroup$
    – CL.
    Dec 4 '18 at 7:41










2




2




$begingroup$
The direct limit is the set of sequences which are eventually zero.
$endgroup$
– Lord Shark the Unknown
Dec 4 '18 at 7:40




$begingroup$
The direct limit is the set of sequences which are eventually zero.
$endgroup$
– Lord Shark the Unknown
Dec 4 '18 at 7:40












$begingroup$
Yes, sorry, that is right. But how about the topology?
$endgroup$
– CL.
Dec 4 '18 at 7:41






$begingroup$
Yes, sorry, that is right. But how about the topology?
$endgroup$
– CL.
Dec 4 '18 at 7:41












1 Answer
1






active

oldest

votes


















1












$begingroup$

As mentioned in comments the direct limit is



$$L={(x_1,x_2,x_3,ldots)inmathbb{R}^{mathbb{N}} | x_i=0text{ eventually}}$$



The direct limit topology on $L$ is the topology coherent with ${mathbb{R}^n | ninmathbb{N}}$ treated as subspaces of $L$ via obvious inclusions.



There's another natural choice for topology on $L$: the one induced by the Euclidean norm (note that the Euclidean norm is well defined on sequences that are eventually $0$).



I remember my surprise when I learned that these two are not the same. The example is as follows: let $v_i$ be a vector with $1/i$ on the $i$-th position and $0$ elsewhere. Then $E={e_i}_{i=1}^infty$ is closed in the direct limit (because $Ecapmathbb{R}^n$ is finite for any $n$) but not closed in the Euclidean norm (because $lVert v_irVert=1/i$ and so the sequence converges to $0$).



In fact it can be shown that the direct limit is not metrizable. In particular this also means that $L$ is not a subspace (even up to homeomorphism) of the product topology (at least a product of metrizable spaces) because countable product of metrizable spaces is metrizable.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I wonder what happens if you use the topology where the closed sets are either the full space, or else (images of) closed subsets of some $mathbb{R}^n$. It seems like that would be the smallest topology that makes each $mathbb{R}^n$ have the subspace topology be the Euclidean topology.
    $endgroup$
    – Daniel Schepler
    Dec 6 '18 at 20:48











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

As mentioned in comments the direct limit is



$$L={(x_1,x_2,x_3,ldots)inmathbb{R}^{mathbb{N}} | x_i=0text{ eventually}}$$



The direct limit topology on $L$ is the topology coherent with ${mathbb{R}^n | ninmathbb{N}}$ treated as subspaces of $L$ via obvious inclusions.



There's another natural choice for topology on $L$: the one induced by the Euclidean norm (note that the Euclidean norm is well defined on sequences that are eventually $0$).



I remember my surprise when I learned that these two are not the same. The example is as follows: let $v_i$ be a vector with $1/i$ on the $i$-th position and $0$ elsewhere. Then $E={e_i}_{i=1}^infty$ is closed in the direct limit (because $Ecapmathbb{R}^n$ is finite for any $n$) but not closed in the Euclidean norm (because $lVert v_irVert=1/i$ and so the sequence converges to $0$).



In fact it can be shown that the direct limit is not metrizable. In particular this also means that $L$ is not a subspace (even up to homeomorphism) of the product topology (at least a product of metrizable spaces) because countable product of metrizable spaces is metrizable.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I wonder what happens if you use the topology where the closed sets are either the full space, or else (images of) closed subsets of some $mathbb{R}^n$. It seems like that would be the smallest topology that makes each $mathbb{R}^n$ have the subspace topology be the Euclidean topology.
    $endgroup$
    – Daniel Schepler
    Dec 6 '18 at 20:48
















1












$begingroup$

As mentioned in comments the direct limit is



$$L={(x_1,x_2,x_3,ldots)inmathbb{R}^{mathbb{N}} | x_i=0text{ eventually}}$$



The direct limit topology on $L$ is the topology coherent with ${mathbb{R}^n | ninmathbb{N}}$ treated as subspaces of $L$ via obvious inclusions.



There's another natural choice for topology on $L$: the one induced by the Euclidean norm (note that the Euclidean norm is well defined on sequences that are eventually $0$).



I remember my surprise when I learned that these two are not the same. The example is as follows: let $v_i$ be a vector with $1/i$ on the $i$-th position and $0$ elsewhere. Then $E={e_i}_{i=1}^infty$ is closed in the direct limit (because $Ecapmathbb{R}^n$ is finite for any $n$) but not closed in the Euclidean norm (because $lVert v_irVert=1/i$ and so the sequence converges to $0$).



In fact it can be shown that the direct limit is not metrizable. In particular this also means that $L$ is not a subspace (even up to homeomorphism) of the product topology (at least a product of metrizable spaces) because countable product of metrizable spaces is metrizable.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I wonder what happens if you use the topology where the closed sets are either the full space, or else (images of) closed subsets of some $mathbb{R}^n$. It seems like that would be the smallest topology that makes each $mathbb{R}^n$ have the subspace topology be the Euclidean topology.
    $endgroup$
    – Daniel Schepler
    Dec 6 '18 at 20:48














1












1








1





$begingroup$

As mentioned in comments the direct limit is



$$L={(x_1,x_2,x_3,ldots)inmathbb{R}^{mathbb{N}} | x_i=0text{ eventually}}$$



The direct limit topology on $L$ is the topology coherent with ${mathbb{R}^n | ninmathbb{N}}$ treated as subspaces of $L$ via obvious inclusions.



There's another natural choice for topology on $L$: the one induced by the Euclidean norm (note that the Euclidean norm is well defined on sequences that are eventually $0$).



I remember my surprise when I learned that these two are not the same. The example is as follows: let $v_i$ be a vector with $1/i$ on the $i$-th position and $0$ elsewhere. Then $E={e_i}_{i=1}^infty$ is closed in the direct limit (because $Ecapmathbb{R}^n$ is finite for any $n$) but not closed in the Euclidean norm (because $lVert v_irVert=1/i$ and so the sequence converges to $0$).



In fact it can be shown that the direct limit is not metrizable. In particular this also means that $L$ is not a subspace (even up to homeomorphism) of the product topology (at least a product of metrizable spaces) because countable product of metrizable spaces is metrizable.






share|cite|improve this answer











$endgroup$



As mentioned in comments the direct limit is



$$L={(x_1,x_2,x_3,ldots)inmathbb{R}^{mathbb{N}} | x_i=0text{ eventually}}$$



The direct limit topology on $L$ is the topology coherent with ${mathbb{R}^n | ninmathbb{N}}$ treated as subspaces of $L$ via obvious inclusions.



There's another natural choice for topology on $L$: the one induced by the Euclidean norm (note that the Euclidean norm is well defined on sequences that are eventually $0$).



I remember my surprise when I learned that these two are not the same. The example is as follows: let $v_i$ be a vector with $1/i$ on the $i$-th position and $0$ elsewhere. Then $E={e_i}_{i=1}^infty$ is closed in the direct limit (because $Ecapmathbb{R}^n$ is finite for any $n$) but not closed in the Euclidean norm (because $lVert v_irVert=1/i$ and so the sequence converges to $0$).



In fact it can be shown that the direct limit is not metrizable. In particular this also means that $L$ is not a subspace (even up to homeomorphism) of the product topology (at least a product of metrizable spaces) because countable product of metrizable spaces is metrizable.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 6 '18 at 20:37

























answered Dec 6 '18 at 11:34









freakishfreakish

12.1k1630




12.1k1630












  • $begingroup$
    I wonder what happens if you use the topology where the closed sets are either the full space, or else (images of) closed subsets of some $mathbb{R}^n$. It seems like that would be the smallest topology that makes each $mathbb{R}^n$ have the subspace topology be the Euclidean topology.
    $endgroup$
    – Daniel Schepler
    Dec 6 '18 at 20:48


















  • $begingroup$
    I wonder what happens if you use the topology where the closed sets are either the full space, or else (images of) closed subsets of some $mathbb{R}^n$. It seems like that would be the smallest topology that makes each $mathbb{R}^n$ have the subspace topology be the Euclidean topology.
    $endgroup$
    – Daniel Schepler
    Dec 6 '18 at 20:48
















$begingroup$
I wonder what happens if you use the topology where the closed sets are either the full space, or else (images of) closed subsets of some $mathbb{R}^n$. It seems like that would be the smallest topology that makes each $mathbb{R}^n$ have the subspace topology be the Euclidean topology.
$endgroup$
– Daniel Schepler
Dec 6 '18 at 20:48




$begingroup$
I wonder what happens if you use the topology where the closed sets are either the full space, or else (images of) closed subsets of some $mathbb{R}^n$. It seems like that would be the smallest topology that makes each $mathbb{R}^n$ have the subspace topology be the Euclidean topology.
$endgroup$
– Daniel Schepler
Dec 6 '18 at 20:48


















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