Identity map between metric spaces continuous or not. [closed]
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How to counter this problem? Is it enough to show pre image of unit ball in some metric is open under another to show the continuity? I am not at all getting the path to proceed. And how to contradict when it's not continuous?
real-analysis functional-analysis continuity metric-spaces
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closed as off-topic by user21820, Adrian Keister, max_zorn, YiFan, Eevee Trainer Mar 29 at 2:15
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, Adrian Keister, max_zorn, YiFan, Eevee Trainer
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
How to counter this problem? Is it enough to show pre image of unit ball in some metric is open under another to show the continuity? I am not at all getting the path to proceed. And how to contradict when it's not continuous?
real-analysis functional-analysis continuity metric-spaces
$endgroup$
closed as off-topic by user21820, Adrian Keister, max_zorn, YiFan, Eevee Trainer Mar 29 at 2:15
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, Adrian Keister, max_zorn, YiFan, Eevee Trainer
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
You can use epsilon-delta methods to prove continuity. Proceed as you would with the standard metric but substitute the appropriate metric to make arguments like $d_2(f,g)ltepsilon$.
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– John Douma
Dec 21 '18 at 4:03
2
$begingroup$
Tip: Questions are not well-received on here whenever the question asker doesn't mention any attempted efforts towards a solution. So always try to mention any attempted work even if it seems embarrassingly incorrect, otherwise other users are less motivated to help.
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– Matt A Pelto
Dec 21 '18 at 4:29
add a comment |
$begingroup$
How to counter this problem? Is it enough to show pre image of unit ball in some metric is open under another to show the continuity? I am not at all getting the path to proceed. And how to contradict when it's not continuous?
real-analysis functional-analysis continuity metric-spaces
$endgroup$
How to counter this problem? Is it enough to show pre image of unit ball in some metric is open under another to show the continuity? I am not at all getting the path to proceed. And how to contradict when it's not continuous?
real-analysis functional-analysis continuity metric-spaces
real-analysis functional-analysis continuity metric-spaces
asked Dec 21 '18 at 3:45
ChakSayantanChakSayantan
492511
492511
closed as off-topic by user21820, Adrian Keister, max_zorn, YiFan, Eevee Trainer Mar 29 at 2:15
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, Adrian Keister, max_zorn, YiFan, Eevee Trainer
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by user21820, Adrian Keister, max_zorn, YiFan, Eevee Trainer Mar 29 at 2:15
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, Adrian Keister, max_zorn, YiFan, Eevee Trainer
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
You can use epsilon-delta methods to prove continuity. Proceed as you would with the standard metric but substitute the appropriate metric to make arguments like $d_2(f,g)ltepsilon$.
$endgroup$
– John Douma
Dec 21 '18 at 4:03
2
$begingroup$
Tip: Questions are not well-received on here whenever the question asker doesn't mention any attempted efforts towards a solution. So always try to mention any attempted work even if it seems embarrassingly incorrect, otherwise other users are less motivated to help.
$endgroup$
– Matt A Pelto
Dec 21 '18 at 4:29
add a comment |
$begingroup$
You can use epsilon-delta methods to prove continuity. Proceed as you would with the standard metric but substitute the appropriate metric to make arguments like $d_2(f,g)ltepsilon$.
$endgroup$
– John Douma
Dec 21 '18 at 4:03
2
$begingroup$
Tip: Questions are not well-received on here whenever the question asker doesn't mention any attempted efforts towards a solution. So always try to mention any attempted work even if it seems embarrassingly incorrect, otherwise other users are less motivated to help.
$endgroup$
– Matt A Pelto
Dec 21 '18 at 4:29
$begingroup$
You can use epsilon-delta methods to prove continuity. Proceed as you would with the standard metric but substitute the appropriate metric to make arguments like $d_2(f,g)ltepsilon$.
$endgroup$
– John Douma
Dec 21 '18 at 4:03
$begingroup$
You can use epsilon-delta methods to prove continuity. Proceed as you would with the standard metric but substitute the appropriate metric to make arguments like $d_2(f,g)ltepsilon$.
$endgroup$
– John Douma
Dec 21 '18 at 4:03
2
2
$begingroup$
Tip: Questions are not well-received on here whenever the question asker doesn't mention any attempted efforts towards a solution. So always try to mention any attempted work even if it seems embarrassingly incorrect, otherwise other users are less motivated to help.
$endgroup$
– Matt A Pelto
Dec 21 '18 at 4:29
$begingroup$
Tip: Questions are not well-received on here whenever the question asker doesn't mention any attempted efforts towards a solution. So always try to mention any attempted work even if it seems embarrassingly incorrect, otherwise other users are less motivated to help.
$endgroup$
– Matt A Pelto
Dec 21 '18 at 4:29
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
a. Let $varepsilon>0$ be given. Since $sup_{xin [0,1]} | , f(x)-g(x)| geq int_0^1 |, f(x)-g(x)|dx$, we may choose $delta=varepsilon$ so that $int_0^1 |, f(x)-g(x)|dx<varepsilon$ whenever $sup_{xin [0,1]} |, f(x)-g(x)|<delta$. Therefore $id:X_1 longrightarrow X_2$ is continuous.
b. Notice $int_0^1 |x^n|dx to 0$ as $n to infty$ BUT $sup_{xin [0,1]} |x^n|=1$ for every $n in mathbb N$. Therefore $id:X_2 longrightarrow X_1$ is NOT continuous at the constant function $g equiv 0$ because it is not sequentially continuous there. There are other counterexamples but this is the preferred (by at least 2 people in this thread).
c. I leave to you. The notion of sequential continuity seems good for demonstrating a counterexample to such statements, while the standard notion of continuity seems good for proving such statements true which is not exactly strict advice.
Technically for b. I used the fact that the functions $f_n(x):=x^n$ from the sequence of functions ${f_n}_{n=1}^infty$ converge to the constant function $g equiv 0$ in $(X_2, d_2)$ but the same is not true in $(X_1, d_1)$. And so $id: X_2 longrightarrow X_1$ is not sequentially continuous at $g$ as opposed to identifying the pointwise limit $f(x):=lim_{nto infty} f_n(x)$ which is not in the set $mathcal C[0,1]$ as $f$ is not continuous at $x=1$.
If this seems like an issue, then showing that the chosen sequence of functions is not uniformly Cauchy on $[0,1]$ (Cauchy in $(X_1, d_1)$) avoids this. For this argument we make the following two observations, $|x^n-x^m|=|x^n||1-x^{m-n}|$ and the function $F(x):=left(1-frac1xright)^x$ is nondecreasing on $[1, infty)$ with $lim_{x to infty} F(x)=frac1e$. So with $varepsilon=frac1{4}(1-e^{-1})$ and for any $ngeq 2$, we may select $m=2n$ and have
$$sup_{xin[0,1]} |x^n-x^m| geq left(1-frac1nright)^nleft(1-left(1-frac1nright)^nright)geq varepsilon$$
which shows that the sequence of functions ${f_n}_{n=1}^infty$ is not Cauchy in $(X_1, d_1)$ by definition.
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1
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For the part c, Cauchy Schwartz says, integral 0 to 1 of h < (int 0 to 1 of h²)^1/2. Hence like part a, choosing delta same as given epsilon we get d3 < delta implies d2< epsilon. Therefore id: X3 to X2 is continuous.
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– ChakSayantan
Dec 21 '18 at 6:09
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You got it, btw I didn't downvote your question. Someone else did and so I shared my tip. My upvote was preemptive but I would say you proved worthy. Most people seem eager to rush towards negative judgement these days (not to say some don't deserve it...rhymes with rump).
$endgroup$
– Matt A Pelto
Dec 21 '18 at 9:43
$begingroup$
Oh but learn to write in latex too. Tip: right click>"show math as" AND for symbols there are pdf documents littering the web with thorough latex code catalogs.
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– Matt A Pelto
Dec 21 '18 at 10:15
1
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Regarding $a.$ If $d_1,d_2$ are metrics on a set $X$ and if there exists $K>0$ such that $d_2(f,g)leq Kcdot d_1(f,g)$ for all $f,gin X$ then the topology generated by $d_2$ is a subset of the topology generated by $d_1,$ so $ id_X:(X,d_1)to (X,d_2)$ is continuous because the inverse of a $d_2$-open set, which is itself, is also $d_1$-open.... In $a.$ we have $K=1.$
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– DanielWainfleet
Dec 22 '18 at 3:06
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Comparing topologies on the same set certainly can come in handy sometimes: imgur.com/cfOxqfs (an old assignment I did that plays off the same notion). I guess considering a and b together, we might say that an open bijection is not always continuous -_-
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– Matt A Pelto
Dec 22 '18 at 3:24
add a comment |
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We know that a continuous map between metric spaces is one which preserves limits: $f$ is continuous if and only if $limlimits_{ntoinfty} f(x_n)=f(limlimits_{ntoinfty} x_n)$ for each convergent sequence $x_n$. To prove that a map is not continuous, we can try to find a sequence that is convergent in one metric but not the other.
For example, I don't think that b) is true. Consider the sequence
$$f_n(x)=x^n.$$
Then $limlimits_{ntoinfty} f_n=0$ in the metric $d_2$, but not in $d_1$: it even fails to be Cauchy in $d_1$.
Of course, in general it is fine to show that the preimage of any open ball is open. There is some general theory that tells you that if the identity map is bounded - that is, if for every $xin X_i$, $d_j(x,0)le Mcdot d_i(x,0)$ -- then the identity map from $X_i$ to $X_j$ is continuous. This uses the fact that all these metrics come from norms, though, and takes some work to establish.
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nice counterexample
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– Matt A Pelto
Dec 21 '18 at 4:25
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
a. Let $varepsilon>0$ be given. Since $sup_{xin [0,1]} | , f(x)-g(x)| geq int_0^1 |, f(x)-g(x)|dx$, we may choose $delta=varepsilon$ so that $int_0^1 |, f(x)-g(x)|dx<varepsilon$ whenever $sup_{xin [0,1]} |, f(x)-g(x)|<delta$. Therefore $id:X_1 longrightarrow X_2$ is continuous.
b. Notice $int_0^1 |x^n|dx to 0$ as $n to infty$ BUT $sup_{xin [0,1]} |x^n|=1$ for every $n in mathbb N$. Therefore $id:X_2 longrightarrow X_1$ is NOT continuous at the constant function $g equiv 0$ because it is not sequentially continuous there. There are other counterexamples but this is the preferred (by at least 2 people in this thread).
c. I leave to you. The notion of sequential continuity seems good for demonstrating a counterexample to such statements, while the standard notion of continuity seems good for proving such statements true which is not exactly strict advice.
Technically for b. I used the fact that the functions $f_n(x):=x^n$ from the sequence of functions ${f_n}_{n=1}^infty$ converge to the constant function $g equiv 0$ in $(X_2, d_2)$ but the same is not true in $(X_1, d_1)$. And so $id: X_2 longrightarrow X_1$ is not sequentially continuous at $g$ as opposed to identifying the pointwise limit $f(x):=lim_{nto infty} f_n(x)$ which is not in the set $mathcal C[0,1]$ as $f$ is not continuous at $x=1$.
If this seems like an issue, then showing that the chosen sequence of functions is not uniformly Cauchy on $[0,1]$ (Cauchy in $(X_1, d_1)$) avoids this. For this argument we make the following two observations, $|x^n-x^m|=|x^n||1-x^{m-n}|$ and the function $F(x):=left(1-frac1xright)^x$ is nondecreasing on $[1, infty)$ with $lim_{x to infty} F(x)=frac1e$. So with $varepsilon=frac1{4}(1-e^{-1})$ and for any $ngeq 2$, we may select $m=2n$ and have
$$sup_{xin[0,1]} |x^n-x^m| geq left(1-frac1nright)^nleft(1-left(1-frac1nright)^nright)geq varepsilon$$
which shows that the sequence of functions ${f_n}_{n=1}^infty$ is not Cauchy in $(X_1, d_1)$ by definition.
$endgroup$
1
$begingroup$
For the part c, Cauchy Schwartz says, integral 0 to 1 of h < (int 0 to 1 of h²)^1/2. Hence like part a, choosing delta same as given epsilon we get d3 < delta implies d2< epsilon. Therefore id: X3 to X2 is continuous.
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– ChakSayantan
Dec 21 '18 at 6:09
$begingroup$
You got it, btw I didn't downvote your question. Someone else did and so I shared my tip. My upvote was preemptive but I would say you proved worthy. Most people seem eager to rush towards negative judgement these days (not to say some don't deserve it...rhymes with rump).
$endgroup$
– Matt A Pelto
Dec 21 '18 at 9:43
$begingroup$
Oh but learn to write in latex too. Tip: right click>"show math as" AND for symbols there are pdf documents littering the web with thorough latex code catalogs.
$endgroup$
– Matt A Pelto
Dec 21 '18 at 10:15
1
$begingroup$
Regarding $a.$ If $d_1,d_2$ are metrics on a set $X$ and if there exists $K>0$ such that $d_2(f,g)leq Kcdot d_1(f,g)$ for all $f,gin X$ then the topology generated by $d_2$ is a subset of the topology generated by $d_1,$ so $ id_X:(X,d_1)to (X,d_2)$ is continuous because the inverse of a $d_2$-open set, which is itself, is also $d_1$-open.... In $a.$ we have $K=1.$
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– DanielWainfleet
Dec 22 '18 at 3:06
$begingroup$
Comparing topologies on the same set certainly can come in handy sometimes: imgur.com/cfOxqfs (an old assignment I did that plays off the same notion). I guess considering a and b together, we might say that an open bijection is not always continuous -_-
$endgroup$
– Matt A Pelto
Dec 22 '18 at 3:24
add a comment |
$begingroup$
a. Let $varepsilon>0$ be given. Since $sup_{xin [0,1]} | , f(x)-g(x)| geq int_0^1 |, f(x)-g(x)|dx$, we may choose $delta=varepsilon$ so that $int_0^1 |, f(x)-g(x)|dx<varepsilon$ whenever $sup_{xin [0,1]} |, f(x)-g(x)|<delta$. Therefore $id:X_1 longrightarrow X_2$ is continuous.
b. Notice $int_0^1 |x^n|dx to 0$ as $n to infty$ BUT $sup_{xin [0,1]} |x^n|=1$ for every $n in mathbb N$. Therefore $id:X_2 longrightarrow X_1$ is NOT continuous at the constant function $g equiv 0$ because it is not sequentially continuous there. There are other counterexamples but this is the preferred (by at least 2 people in this thread).
c. I leave to you. The notion of sequential continuity seems good for demonstrating a counterexample to such statements, while the standard notion of continuity seems good for proving such statements true which is not exactly strict advice.
Technically for b. I used the fact that the functions $f_n(x):=x^n$ from the sequence of functions ${f_n}_{n=1}^infty$ converge to the constant function $g equiv 0$ in $(X_2, d_2)$ but the same is not true in $(X_1, d_1)$. And so $id: X_2 longrightarrow X_1$ is not sequentially continuous at $g$ as opposed to identifying the pointwise limit $f(x):=lim_{nto infty} f_n(x)$ which is not in the set $mathcal C[0,1]$ as $f$ is not continuous at $x=1$.
If this seems like an issue, then showing that the chosen sequence of functions is not uniformly Cauchy on $[0,1]$ (Cauchy in $(X_1, d_1)$) avoids this. For this argument we make the following two observations, $|x^n-x^m|=|x^n||1-x^{m-n}|$ and the function $F(x):=left(1-frac1xright)^x$ is nondecreasing on $[1, infty)$ with $lim_{x to infty} F(x)=frac1e$. So with $varepsilon=frac1{4}(1-e^{-1})$ and for any $ngeq 2$, we may select $m=2n$ and have
$$sup_{xin[0,1]} |x^n-x^m| geq left(1-frac1nright)^nleft(1-left(1-frac1nright)^nright)geq varepsilon$$
which shows that the sequence of functions ${f_n}_{n=1}^infty$ is not Cauchy in $(X_1, d_1)$ by definition.
$endgroup$
1
$begingroup$
For the part c, Cauchy Schwartz says, integral 0 to 1 of h < (int 0 to 1 of h²)^1/2. Hence like part a, choosing delta same as given epsilon we get d3 < delta implies d2< epsilon. Therefore id: X3 to X2 is continuous.
$endgroup$
– ChakSayantan
Dec 21 '18 at 6:09
$begingroup$
You got it, btw I didn't downvote your question. Someone else did and so I shared my tip. My upvote was preemptive but I would say you proved worthy. Most people seem eager to rush towards negative judgement these days (not to say some don't deserve it...rhymes with rump).
$endgroup$
– Matt A Pelto
Dec 21 '18 at 9:43
$begingroup$
Oh but learn to write in latex too. Tip: right click>"show math as" AND for symbols there are pdf documents littering the web with thorough latex code catalogs.
$endgroup$
– Matt A Pelto
Dec 21 '18 at 10:15
1
$begingroup$
Regarding $a.$ If $d_1,d_2$ are metrics on a set $X$ and if there exists $K>0$ such that $d_2(f,g)leq Kcdot d_1(f,g)$ for all $f,gin X$ then the topology generated by $d_2$ is a subset of the topology generated by $d_1,$ so $ id_X:(X,d_1)to (X,d_2)$ is continuous because the inverse of a $d_2$-open set, which is itself, is also $d_1$-open.... In $a.$ we have $K=1.$
$endgroup$
– DanielWainfleet
Dec 22 '18 at 3:06
$begingroup$
Comparing topologies on the same set certainly can come in handy sometimes: imgur.com/cfOxqfs (an old assignment I did that plays off the same notion). I guess considering a and b together, we might say that an open bijection is not always continuous -_-
$endgroup$
– Matt A Pelto
Dec 22 '18 at 3:24
add a comment |
$begingroup$
a. Let $varepsilon>0$ be given. Since $sup_{xin [0,1]} | , f(x)-g(x)| geq int_0^1 |, f(x)-g(x)|dx$, we may choose $delta=varepsilon$ so that $int_0^1 |, f(x)-g(x)|dx<varepsilon$ whenever $sup_{xin [0,1]} |, f(x)-g(x)|<delta$. Therefore $id:X_1 longrightarrow X_2$ is continuous.
b. Notice $int_0^1 |x^n|dx to 0$ as $n to infty$ BUT $sup_{xin [0,1]} |x^n|=1$ for every $n in mathbb N$. Therefore $id:X_2 longrightarrow X_1$ is NOT continuous at the constant function $g equiv 0$ because it is not sequentially continuous there. There are other counterexamples but this is the preferred (by at least 2 people in this thread).
c. I leave to you. The notion of sequential continuity seems good for demonstrating a counterexample to such statements, while the standard notion of continuity seems good for proving such statements true which is not exactly strict advice.
Technically for b. I used the fact that the functions $f_n(x):=x^n$ from the sequence of functions ${f_n}_{n=1}^infty$ converge to the constant function $g equiv 0$ in $(X_2, d_2)$ but the same is not true in $(X_1, d_1)$. And so $id: X_2 longrightarrow X_1$ is not sequentially continuous at $g$ as opposed to identifying the pointwise limit $f(x):=lim_{nto infty} f_n(x)$ which is not in the set $mathcal C[0,1]$ as $f$ is not continuous at $x=1$.
If this seems like an issue, then showing that the chosen sequence of functions is not uniformly Cauchy on $[0,1]$ (Cauchy in $(X_1, d_1)$) avoids this. For this argument we make the following two observations, $|x^n-x^m|=|x^n||1-x^{m-n}|$ and the function $F(x):=left(1-frac1xright)^x$ is nondecreasing on $[1, infty)$ with $lim_{x to infty} F(x)=frac1e$. So with $varepsilon=frac1{4}(1-e^{-1})$ and for any $ngeq 2$, we may select $m=2n$ and have
$$sup_{xin[0,1]} |x^n-x^m| geq left(1-frac1nright)^nleft(1-left(1-frac1nright)^nright)geq varepsilon$$
which shows that the sequence of functions ${f_n}_{n=1}^infty$ is not Cauchy in $(X_1, d_1)$ by definition.
$endgroup$
a. Let $varepsilon>0$ be given. Since $sup_{xin [0,1]} | , f(x)-g(x)| geq int_0^1 |, f(x)-g(x)|dx$, we may choose $delta=varepsilon$ so that $int_0^1 |, f(x)-g(x)|dx<varepsilon$ whenever $sup_{xin [0,1]} |, f(x)-g(x)|<delta$. Therefore $id:X_1 longrightarrow X_2$ is continuous.
b. Notice $int_0^1 |x^n|dx to 0$ as $n to infty$ BUT $sup_{xin [0,1]} |x^n|=1$ for every $n in mathbb N$. Therefore $id:X_2 longrightarrow X_1$ is NOT continuous at the constant function $g equiv 0$ because it is not sequentially continuous there. There are other counterexamples but this is the preferred (by at least 2 people in this thread).
c. I leave to you. The notion of sequential continuity seems good for demonstrating a counterexample to such statements, while the standard notion of continuity seems good for proving such statements true which is not exactly strict advice.
Technically for b. I used the fact that the functions $f_n(x):=x^n$ from the sequence of functions ${f_n}_{n=1}^infty$ converge to the constant function $g equiv 0$ in $(X_2, d_2)$ but the same is not true in $(X_1, d_1)$. And so $id: X_2 longrightarrow X_1$ is not sequentially continuous at $g$ as opposed to identifying the pointwise limit $f(x):=lim_{nto infty} f_n(x)$ which is not in the set $mathcal C[0,1]$ as $f$ is not continuous at $x=1$.
If this seems like an issue, then showing that the chosen sequence of functions is not uniformly Cauchy on $[0,1]$ (Cauchy in $(X_1, d_1)$) avoids this. For this argument we make the following two observations, $|x^n-x^m|=|x^n||1-x^{m-n}|$ and the function $F(x):=left(1-frac1xright)^x$ is nondecreasing on $[1, infty)$ with $lim_{x to infty} F(x)=frac1e$. So with $varepsilon=frac1{4}(1-e^{-1})$ and for any $ngeq 2$, we may select $m=2n$ and have
$$sup_{xin[0,1]} |x^n-x^m| geq left(1-frac1nright)^nleft(1-left(1-frac1nright)^nright)geq varepsilon$$
which shows that the sequence of functions ${f_n}_{n=1}^infty$ is not Cauchy in $(X_1, d_1)$ by definition.
edited Dec 23 '18 at 20:10
answered Dec 21 '18 at 4:23
Matt A PeltoMatt A Pelto
2,677621
2,677621
1
$begingroup$
For the part c, Cauchy Schwartz says, integral 0 to 1 of h < (int 0 to 1 of h²)^1/2. Hence like part a, choosing delta same as given epsilon we get d3 < delta implies d2< epsilon. Therefore id: X3 to X2 is continuous.
$endgroup$
– ChakSayantan
Dec 21 '18 at 6:09
$begingroup$
You got it, btw I didn't downvote your question. Someone else did and so I shared my tip. My upvote was preemptive but I would say you proved worthy. Most people seem eager to rush towards negative judgement these days (not to say some don't deserve it...rhymes with rump).
$endgroup$
– Matt A Pelto
Dec 21 '18 at 9:43
$begingroup$
Oh but learn to write in latex too. Tip: right click>"show math as" AND for symbols there are pdf documents littering the web with thorough latex code catalogs.
$endgroup$
– Matt A Pelto
Dec 21 '18 at 10:15
1
$begingroup$
Regarding $a.$ If $d_1,d_2$ are metrics on a set $X$ and if there exists $K>0$ such that $d_2(f,g)leq Kcdot d_1(f,g)$ for all $f,gin X$ then the topology generated by $d_2$ is a subset of the topology generated by $d_1,$ so $ id_X:(X,d_1)to (X,d_2)$ is continuous because the inverse of a $d_2$-open set, which is itself, is also $d_1$-open.... In $a.$ we have $K=1.$
$endgroup$
– DanielWainfleet
Dec 22 '18 at 3:06
$begingroup$
Comparing topologies on the same set certainly can come in handy sometimes: imgur.com/cfOxqfs (an old assignment I did that plays off the same notion). I guess considering a and b together, we might say that an open bijection is not always continuous -_-
$endgroup$
– Matt A Pelto
Dec 22 '18 at 3:24
add a comment |
1
$begingroup$
For the part c, Cauchy Schwartz says, integral 0 to 1 of h < (int 0 to 1 of h²)^1/2. Hence like part a, choosing delta same as given epsilon we get d3 < delta implies d2< epsilon. Therefore id: X3 to X2 is continuous.
$endgroup$
– ChakSayantan
Dec 21 '18 at 6:09
$begingroup$
You got it, btw I didn't downvote your question. Someone else did and so I shared my tip. My upvote was preemptive but I would say you proved worthy. Most people seem eager to rush towards negative judgement these days (not to say some don't deserve it...rhymes with rump).
$endgroup$
– Matt A Pelto
Dec 21 '18 at 9:43
$begingroup$
Oh but learn to write in latex too. Tip: right click>"show math as" AND for symbols there are pdf documents littering the web with thorough latex code catalogs.
$endgroup$
– Matt A Pelto
Dec 21 '18 at 10:15
1
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Regarding $a.$ If $d_1,d_2$ are metrics on a set $X$ and if there exists $K>0$ such that $d_2(f,g)leq Kcdot d_1(f,g)$ for all $f,gin X$ then the topology generated by $d_2$ is a subset of the topology generated by $d_1,$ so $ id_X:(X,d_1)to (X,d_2)$ is continuous because the inverse of a $d_2$-open set, which is itself, is also $d_1$-open.... In $a.$ we have $K=1.$
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– DanielWainfleet
Dec 22 '18 at 3:06
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Comparing topologies on the same set certainly can come in handy sometimes: imgur.com/cfOxqfs (an old assignment I did that plays off the same notion). I guess considering a and b together, we might say that an open bijection is not always continuous -_-
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– Matt A Pelto
Dec 22 '18 at 3:24
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For the part c, Cauchy Schwartz says, integral 0 to 1 of h < (int 0 to 1 of h²)^1/2. Hence like part a, choosing delta same as given epsilon we get d3 < delta implies d2< epsilon. Therefore id: X3 to X2 is continuous.
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– ChakSayantan
Dec 21 '18 at 6:09
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For the part c, Cauchy Schwartz says, integral 0 to 1 of h < (int 0 to 1 of h²)^1/2. Hence like part a, choosing delta same as given epsilon we get d3 < delta implies d2< epsilon. Therefore id: X3 to X2 is continuous.
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– ChakSayantan
Dec 21 '18 at 6:09
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You got it, btw I didn't downvote your question. Someone else did and so I shared my tip. My upvote was preemptive but I would say you proved worthy. Most people seem eager to rush towards negative judgement these days (not to say some don't deserve it...rhymes with rump).
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– Matt A Pelto
Dec 21 '18 at 9:43
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You got it, btw I didn't downvote your question. Someone else did and so I shared my tip. My upvote was preemptive but I would say you proved worthy. Most people seem eager to rush towards negative judgement these days (not to say some don't deserve it...rhymes with rump).
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– Matt A Pelto
Dec 21 '18 at 9:43
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Oh but learn to write in latex too. Tip: right click>"show math as" AND for symbols there are pdf documents littering the web with thorough latex code catalogs.
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– Matt A Pelto
Dec 21 '18 at 10:15
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Oh but learn to write in latex too. Tip: right click>"show math as" AND for symbols there are pdf documents littering the web with thorough latex code catalogs.
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– Matt A Pelto
Dec 21 '18 at 10:15
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Regarding $a.$ If $d_1,d_2$ are metrics on a set $X$ and if there exists $K>0$ such that $d_2(f,g)leq Kcdot d_1(f,g)$ for all $f,gin X$ then the topology generated by $d_2$ is a subset of the topology generated by $d_1,$ so $ id_X:(X,d_1)to (X,d_2)$ is continuous because the inverse of a $d_2$-open set, which is itself, is also $d_1$-open.... In $a.$ we have $K=1.$
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– DanielWainfleet
Dec 22 '18 at 3:06
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Regarding $a.$ If $d_1,d_2$ are metrics on a set $X$ and if there exists $K>0$ such that $d_2(f,g)leq Kcdot d_1(f,g)$ for all $f,gin X$ then the topology generated by $d_2$ is a subset of the topology generated by $d_1,$ so $ id_X:(X,d_1)to (X,d_2)$ is continuous because the inverse of a $d_2$-open set, which is itself, is also $d_1$-open.... In $a.$ we have $K=1.$
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– DanielWainfleet
Dec 22 '18 at 3:06
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Comparing topologies on the same set certainly can come in handy sometimes: imgur.com/cfOxqfs (an old assignment I did that plays off the same notion). I guess considering a and b together, we might say that an open bijection is not always continuous -_-
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– Matt A Pelto
Dec 22 '18 at 3:24
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Comparing topologies on the same set certainly can come in handy sometimes: imgur.com/cfOxqfs (an old assignment I did that plays off the same notion). I guess considering a and b together, we might say that an open bijection is not always continuous -_-
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– Matt A Pelto
Dec 22 '18 at 3:24
add a comment |
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We know that a continuous map between metric spaces is one which preserves limits: $f$ is continuous if and only if $limlimits_{ntoinfty} f(x_n)=f(limlimits_{ntoinfty} x_n)$ for each convergent sequence $x_n$. To prove that a map is not continuous, we can try to find a sequence that is convergent in one metric but not the other.
For example, I don't think that b) is true. Consider the sequence
$$f_n(x)=x^n.$$
Then $limlimits_{ntoinfty} f_n=0$ in the metric $d_2$, but not in $d_1$: it even fails to be Cauchy in $d_1$.
Of course, in general it is fine to show that the preimage of any open ball is open. There is some general theory that tells you that if the identity map is bounded - that is, if for every $xin X_i$, $d_j(x,0)le Mcdot d_i(x,0)$ -- then the identity map from $X_i$ to $X_j$ is continuous. This uses the fact that all these metrics come from norms, though, and takes some work to establish.
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nice counterexample
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– Matt A Pelto
Dec 21 '18 at 4:25
add a comment |
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We know that a continuous map between metric spaces is one which preserves limits: $f$ is continuous if and only if $limlimits_{ntoinfty} f(x_n)=f(limlimits_{ntoinfty} x_n)$ for each convergent sequence $x_n$. To prove that a map is not continuous, we can try to find a sequence that is convergent in one metric but not the other.
For example, I don't think that b) is true. Consider the sequence
$$f_n(x)=x^n.$$
Then $limlimits_{ntoinfty} f_n=0$ in the metric $d_2$, but not in $d_1$: it even fails to be Cauchy in $d_1$.
Of course, in general it is fine to show that the preimage of any open ball is open. There is some general theory that tells you that if the identity map is bounded - that is, if for every $xin X_i$, $d_j(x,0)le Mcdot d_i(x,0)$ -- then the identity map from $X_i$ to $X_j$ is continuous. This uses the fact that all these metrics come from norms, though, and takes some work to establish.
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nice counterexample
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– Matt A Pelto
Dec 21 '18 at 4:25
add a comment |
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We know that a continuous map between metric spaces is one which preserves limits: $f$ is continuous if and only if $limlimits_{ntoinfty} f(x_n)=f(limlimits_{ntoinfty} x_n)$ for each convergent sequence $x_n$. To prove that a map is not continuous, we can try to find a sequence that is convergent in one metric but not the other.
For example, I don't think that b) is true. Consider the sequence
$$f_n(x)=x^n.$$
Then $limlimits_{ntoinfty} f_n=0$ in the metric $d_2$, but not in $d_1$: it even fails to be Cauchy in $d_1$.
Of course, in general it is fine to show that the preimage of any open ball is open. There is some general theory that tells you that if the identity map is bounded - that is, if for every $xin X_i$, $d_j(x,0)le Mcdot d_i(x,0)$ -- then the identity map from $X_i$ to $X_j$ is continuous. This uses the fact that all these metrics come from norms, though, and takes some work to establish.
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We know that a continuous map between metric spaces is one which preserves limits: $f$ is continuous if and only if $limlimits_{ntoinfty} f(x_n)=f(limlimits_{ntoinfty} x_n)$ for each convergent sequence $x_n$. To prove that a map is not continuous, we can try to find a sequence that is convergent in one metric but not the other.
For example, I don't think that b) is true. Consider the sequence
$$f_n(x)=x^n.$$
Then $limlimits_{ntoinfty} f_n=0$ in the metric $d_2$, but not in $d_1$: it even fails to be Cauchy in $d_1$.
Of course, in general it is fine to show that the preimage of any open ball is open. There is some general theory that tells you that if the identity map is bounded - that is, if for every $xin X_i$, $d_j(x,0)le Mcdot d_i(x,0)$ -- then the identity map from $X_i$ to $X_j$ is continuous. This uses the fact that all these metrics come from norms, though, and takes some work to establish.
answered Dec 21 '18 at 3:53
Ashwin TrisalAshwin Trisal
1,2891516
1,2891516
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nice counterexample
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– Matt A Pelto
Dec 21 '18 at 4:25
add a comment |
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nice counterexample
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– Matt A Pelto
Dec 21 '18 at 4:25
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nice counterexample
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– Matt A Pelto
Dec 21 '18 at 4:25
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nice counterexample
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– Matt A Pelto
Dec 21 '18 at 4:25
add a comment |
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You can use epsilon-delta methods to prove continuity. Proceed as you would with the standard metric but substitute the appropriate metric to make arguments like $d_2(f,g)ltepsilon$.
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– John Douma
Dec 21 '18 at 4:03
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Tip: Questions are not well-received on here whenever the question asker doesn't mention any attempted efforts towards a solution. So always try to mention any attempted work even if it seems embarrassingly incorrect, otherwise other users are less motivated to help.
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– Matt A Pelto
Dec 21 '18 at 4:29