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compactness of a set where am I going wrong
In a metric space, compactness implies completnessClosure and compactness of the set of real eigenvalues of a real matrix.Proving the compactness of a subset of a compact setThe union of finitely many compact subsets of $mathbbR^n$ must be compact.A question about local compactness and $sigma$-compactnessWeakly compactness of a set of finitely additive measuresIf $Y$ is closed subspace of $(X,|cdot|)$, then subspace $<Ycup x_0>$ is closed in $X$.“Decreasing” sequence of nonempty compact sets has a nonempty intersection.Show that $F = lambdain mathbbR_+, xin K$ is closed where $K$ is a compact set.Closed & boundedness, sequentially compactness and Completeness
$begingroup$
I have a proof of the following false fact :
Let $E$ be normed vector space. Let $K subset E$ be a compact set. Then the set $B = lambda x mid lambda in mathbbR^+, x in K $ is closed (where $mathbbR^+$ are the positive real numbers including $0$).
This fact is true when $0 not in K$ yet it can be false when $0 in K$. For example by taking the semi-circle in the plane centered at $(1,0)$ of radius $1$.
So I made a proof a of this fact. My proof is thus obviously false yet I don't see where the mistake is :
Let $(lambda_n k_n)$ be a sequence in $B$ which converges to a vector $x in E$. We want to prove that $x in B$.
Since $K$ is compact there is $phi : mathbbN to mathbbN$ strictly increasing such that $(k_phi(n))$ converges to a vector $k in K$. If $ k = 0$ then the sequence $(lambda_n k_n)$ converges to $0 in B$ and we are done. So we can suppose $k ne 0$.
Since the sequence $(lambda_n k_n)$ converges to $x$ we must have $k in span x $. So there is $mu in mathbbR^*$ such that $k = mu x$. From here we can deduce that the sequence $(lambda_n)$ necessarily converges to $frac1mu$. Yet since $mathbbR^+$ is closed the sequence $(lambda_n)$ converges to a positive real number, so $frac1mu geq 0$ so $mu geq 0$.
So the sequence $(lambda_n k_n)$ converges to the vector $frac1mu k in B$ since $k in K$ and $frac1mu geq 0$. Hence $B$ is closed.
So where am I going wrong here ?
Thank you !
real-analysis general-topology proof-verification compactness
New contributor
$endgroup$
add a comment |
$begingroup$
I have a proof of the following false fact :
Let $E$ be normed vector space. Let $K subset E$ be a compact set. Then the set $B = lambda x mid lambda in mathbbR^+, x in K $ is closed (where $mathbbR^+$ are the positive real numbers including $0$).
This fact is true when $0 not in K$ yet it can be false when $0 in K$. For example by taking the semi-circle in the plane centered at $(1,0)$ of radius $1$.
So I made a proof a of this fact. My proof is thus obviously false yet I don't see where the mistake is :
Let $(lambda_n k_n)$ be a sequence in $B$ which converges to a vector $x in E$. We want to prove that $x in B$.
Since $K$ is compact there is $phi : mathbbN to mathbbN$ strictly increasing such that $(k_phi(n))$ converges to a vector $k in K$. If $ k = 0$ then the sequence $(lambda_n k_n)$ converges to $0 in B$ and we are done. So we can suppose $k ne 0$.
Since the sequence $(lambda_n k_n)$ converges to $x$ we must have $k in span x $. So there is $mu in mathbbR^*$ such that $k = mu x$. From here we can deduce that the sequence $(lambda_n)$ necessarily converges to $frac1mu$. Yet since $mathbbR^+$ is closed the sequence $(lambda_n)$ converges to a positive real number, so $frac1mu geq 0$ so $mu geq 0$.
So the sequence $(lambda_n k_n)$ converges to the vector $frac1mu k in B$ since $k in K$ and $frac1mu geq 0$. Hence $B$ is closed.
So where am I going wrong here ?
Thank you !
real-analysis general-topology proof-verification compactness
New contributor
$endgroup$
$begingroup$
I guess you mean $B$ to be the union of the $lambda K$ rather than the set of them?
$endgroup$
– Eric Wofsey
6 hours ago
$begingroup$
@EricWofsey By this notation I mean the set of $lambda x$ for all $x in K$ and $lambda geq 0$, so my notation is wrong ?
$endgroup$
– mouargmouarg
6 hours ago
$begingroup$
Yes, your notation meant that $B$ is the set whose elements are the sets $lambda K$ (so an element of $B$ is a subset of $E$, not a point of $E$).
$endgroup$
– Eric Wofsey
6 hours ago
$begingroup$
@EricWofsey You are right thank you.
$endgroup$
– mouargmouarg
6 hours ago
$begingroup$
Note that you state that $mathbbR^+$ is closed, which isn’t the case
$endgroup$
– Santana Afton
51 mins ago
add a comment |
$begingroup$
I have a proof of the following false fact :
Let $E$ be normed vector space. Let $K subset E$ be a compact set. Then the set $B = lambda x mid lambda in mathbbR^+, x in K $ is closed (where $mathbbR^+$ are the positive real numbers including $0$).
This fact is true when $0 not in K$ yet it can be false when $0 in K$. For example by taking the semi-circle in the plane centered at $(1,0)$ of radius $1$.
So I made a proof a of this fact. My proof is thus obviously false yet I don't see where the mistake is :
Let $(lambda_n k_n)$ be a sequence in $B$ which converges to a vector $x in E$. We want to prove that $x in B$.
Since $K$ is compact there is $phi : mathbbN to mathbbN$ strictly increasing such that $(k_phi(n))$ converges to a vector $k in K$. If $ k = 0$ then the sequence $(lambda_n k_n)$ converges to $0 in B$ and we are done. So we can suppose $k ne 0$.
Since the sequence $(lambda_n k_n)$ converges to $x$ we must have $k in span x $. So there is $mu in mathbbR^*$ such that $k = mu x$. From here we can deduce that the sequence $(lambda_n)$ necessarily converges to $frac1mu$. Yet since $mathbbR^+$ is closed the sequence $(lambda_n)$ converges to a positive real number, so $frac1mu geq 0$ so $mu geq 0$.
So the sequence $(lambda_n k_n)$ converges to the vector $frac1mu k in B$ since $k in K$ and $frac1mu geq 0$. Hence $B$ is closed.
So where am I going wrong here ?
Thank you !
real-analysis general-topology proof-verification compactness
New contributor
$endgroup$
I have a proof of the following false fact :
Let $E$ be normed vector space. Let $K subset E$ be a compact set. Then the set $B = lambda x mid lambda in mathbbR^+, x in K $ is closed (where $mathbbR^+$ are the positive real numbers including $0$).
This fact is true when $0 not in K$ yet it can be false when $0 in K$. For example by taking the semi-circle in the plane centered at $(1,0)$ of radius $1$.
So I made a proof a of this fact. My proof is thus obviously false yet I don't see where the mistake is :
Let $(lambda_n k_n)$ be a sequence in $B$ which converges to a vector $x in E$. We want to prove that $x in B$.
Since $K$ is compact there is $phi : mathbbN to mathbbN$ strictly increasing such that $(k_phi(n))$ converges to a vector $k in K$. If $ k = 0$ then the sequence $(lambda_n k_n)$ converges to $0 in B$ and we are done. So we can suppose $k ne 0$.
Since the sequence $(lambda_n k_n)$ converges to $x$ we must have $k in span x $. So there is $mu in mathbbR^*$ such that $k = mu x$. From here we can deduce that the sequence $(lambda_n)$ necessarily converges to $frac1mu$. Yet since $mathbbR^+$ is closed the sequence $(lambda_n)$ converges to a positive real number, so $frac1mu geq 0$ so $mu geq 0$.
So the sequence $(lambda_n k_n)$ converges to the vector $frac1mu k in B$ since $k in K$ and $frac1mu geq 0$. Hence $B$ is closed.
So where am I going wrong here ?
Thank you !
real-analysis general-topology proof-verification compactness
real-analysis general-topology proof-verification compactness
New contributor
New contributor
edited 6 hours ago
mouargmouarg
New contributor
asked 6 hours ago
mouargmouargmouargmouarg
533
533
New contributor
New contributor
$begingroup$
I guess you mean $B$ to be the union of the $lambda K$ rather than the set of them?
$endgroup$
– Eric Wofsey
6 hours ago
$begingroup$
@EricWofsey By this notation I mean the set of $lambda x$ for all $x in K$ and $lambda geq 0$, so my notation is wrong ?
$endgroup$
– mouargmouarg
6 hours ago
$begingroup$
Yes, your notation meant that $B$ is the set whose elements are the sets $lambda K$ (so an element of $B$ is a subset of $E$, not a point of $E$).
$endgroup$
– Eric Wofsey
6 hours ago
$begingroup$
@EricWofsey You are right thank you.
$endgroup$
– mouargmouarg
6 hours ago
$begingroup$
Note that you state that $mathbbR^+$ is closed, which isn’t the case
$endgroup$
– Santana Afton
51 mins ago
add a comment |
$begingroup$
I guess you mean $B$ to be the union of the $lambda K$ rather than the set of them?
$endgroup$
– Eric Wofsey
6 hours ago
$begingroup$
@EricWofsey By this notation I mean the set of $lambda x$ for all $x in K$ and $lambda geq 0$, so my notation is wrong ?
$endgroup$
– mouargmouarg
6 hours ago
$begingroup$
Yes, your notation meant that $B$ is the set whose elements are the sets $lambda K$ (so an element of $B$ is a subset of $E$, not a point of $E$).
$endgroup$
– Eric Wofsey
6 hours ago
$begingroup$
@EricWofsey You are right thank you.
$endgroup$
– mouargmouarg
6 hours ago
$begingroup$
Note that you state that $mathbbR^+$ is closed, which isn’t the case
$endgroup$
– Santana Afton
51 mins ago
$begingroup$
I guess you mean $B$ to be the union of the $lambda K$ rather than the set of them?
$endgroup$
– Eric Wofsey
6 hours ago
$begingroup$
I guess you mean $B$ to be the union of the $lambda K$ rather than the set of them?
$endgroup$
– Eric Wofsey
6 hours ago
$begingroup$
@EricWofsey By this notation I mean the set of $lambda x$ for all $x in K$ and $lambda geq 0$, so my notation is wrong ?
$endgroup$
– mouargmouarg
6 hours ago
$begingroup$
@EricWofsey By this notation I mean the set of $lambda x$ for all $x in K$ and $lambda geq 0$, so my notation is wrong ?
$endgroup$
– mouargmouarg
6 hours ago
$begingroup$
Yes, your notation meant that $B$ is the set whose elements are the sets $lambda K$ (so an element of $B$ is a subset of $E$, not a point of $E$).
$endgroup$
– Eric Wofsey
6 hours ago
$begingroup$
Yes, your notation meant that $B$ is the set whose elements are the sets $lambda K$ (so an element of $B$ is a subset of $E$, not a point of $E$).
$endgroup$
– Eric Wofsey
6 hours ago
$begingroup$
@EricWofsey You are right thank you.
$endgroup$
– mouargmouarg
6 hours ago
$begingroup$
@EricWofsey You are right thank you.
$endgroup$
– mouargmouarg
6 hours ago
$begingroup$
Note that you state that $mathbbR^+$ is closed, which isn’t the case
$endgroup$
– Santana Afton
51 mins ago
$begingroup$
Note that you state that $mathbbR^+$ is closed, which isn’t the case
$endgroup$
– Santana Afton
51 mins ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
In this situation, try to apply your argument to the counter-example you found, to see where it goes wrong.
Let us take for $K$ the circle of center $(1,0)$ and radius $1$. Then $S := bigcup_lambda geq 0 lambda K = (0,0) cup (mathbbR_+^* times mathbbR)$. It is not closed because one can take the family $f(t) = (t,1)$, which belongs to $S$ for $t>0$, and converges to $(0,1) notin S$ as $t$ goes to $0$.
The corresponding family of parameters $lambda (t)$ and $k(t)$ are:
$$lambda(t) = frac1+t^22t, quad k(t) = frac2t1+t^2 (t,1).$$
As $t$ goes to $0$, we have that $k(t)$ converges to $0$. But $lambda (t) k(t)$ does not converge to $0$, because $lambda(t)$ increases fast enough to compensate for the decay of $k(t)$.
hence your mistake is there: the fact that $(k_n)$ converges to $0$ does not imply that $(lambda_n k_n)$ also converges to $0$, because $lambda_n$ has no reason to be bounded.
$endgroup$
$begingroup$
You are right, it's more clear to me what's happening now. Thank you.
$endgroup$
– mouargmouarg
6 hours ago
add a comment |
$begingroup$
If $ k = 0$ then the sequence $(lambda_n k_n)$ converges to $0 in B$ and we are done.
This is wrong. We know $k_phi(n)to 0$, but $lambda_n$ may be getting large so $lambda_nk_n$ can converge to a nonzero value.
$endgroup$
$begingroup$
Oh, thank you. I should have spotted that... For example by taking : $k_n = frac1n$ and $lambda_n = n$
$endgroup$
– mouargmouarg
6 hours ago
add a comment |
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2 Answers
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2 Answers
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$begingroup$
In this situation, try to apply your argument to the counter-example you found, to see where it goes wrong.
Let us take for $K$ the circle of center $(1,0)$ and radius $1$. Then $S := bigcup_lambda geq 0 lambda K = (0,0) cup (mathbbR_+^* times mathbbR)$. It is not closed because one can take the family $f(t) = (t,1)$, which belongs to $S$ for $t>0$, and converges to $(0,1) notin S$ as $t$ goes to $0$.
The corresponding family of parameters $lambda (t)$ and $k(t)$ are:
$$lambda(t) = frac1+t^22t, quad k(t) = frac2t1+t^2 (t,1).$$
As $t$ goes to $0$, we have that $k(t)$ converges to $0$. But $lambda (t) k(t)$ does not converge to $0$, because $lambda(t)$ increases fast enough to compensate for the decay of $k(t)$.
hence your mistake is there: the fact that $(k_n)$ converges to $0$ does not imply that $(lambda_n k_n)$ also converges to $0$, because $lambda_n$ has no reason to be bounded.
$endgroup$
$begingroup$
You are right, it's more clear to me what's happening now. Thank you.
$endgroup$
– mouargmouarg
6 hours ago
add a comment |
$begingroup$
In this situation, try to apply your argument to the counter-example you found, to see where it goes wrong.
Let us take for $K$ the circle of center $(1,0)$ and radius $1$. Then $S := bigcup_lambda geq 0 lambda K = (0,0) cup (mathbbR_+^* times mathbbR)$. It is not closed because one can take the family $f(t) = (t,1)$, which belongs to $S$ for $t>0$, and converges to $(0,1) notin S$ as $t$ goes to $0$.
The corresponding family of parameters $lambda (t)$ and $k(t)$ are:
$$lambda(t) = frac1+t^22t, quad k(t) = frac2t1+t^2 (t,1).$$
As $t$ goes to $0$, we have that $k(t)$ converges to $0$. But $lambda (t) k(t)$ does not converge to $0$, because $lambda(t)$ increases fast enough to compensate for the decay of $k(t)$.
hence your mistake is there: the fact that $(k_n)$ converges to $0$ does not imply that $(lambda_n k_n)$ also converges to $0$, because $lambda_n$ has no reason to be bounded.
$endgroup$
$begingroup$
You are right, it's more clear to me what's happening now. Thank you.
$endgroup$
– mouargmouarg
6 hours ago
add a comment |
$begingroup$
In this situation, try to apply your argument to the counter-example you found, to see where it goes wrong.
Let us take for $K$ the circle of center $(1,0)$ and radius $1$. Then $S := bigcup_lambda geq 0 lambda K = (0,0) cup (mathbbR_+^* times mathbbR)$. It is not closed because one can take the family $f(t) = (t,1)$, which belongs to $S$ for $t>0$, and converges to $(0,1) notin S$ as $t$ goes to $0$.
The corresponding family of parameters $lambda (t)$ and $k(t)$ are:
$$lambda(t) = frac1+t^22t, quad k(t) = frac2t1+t^2 (t,1).$$
As $t$ goes to $0$, we have that $k(t)$ converges to $0$. But $lambda (t) k(t)$ does not converge to $0$, because $lambda(t)$ increases fast enough to compensate for the decay of $k(t)$.
hence your mistake is there: the fact that $(k_n)$ converges to $0$ does not imply that $(lambda_n k_n)$ also converges to $0$, because $lambda_n$ has no reason to be bounded.
$endgroup$
In this situation, try to apply your argument to the counter-example you found, to see where it goes wrong.
Let us take for $K$ the circle of center $(1,0)$ and radius $1$. Then $S := bigcup_lambda geq 0 lambda K = (0,0) cup (mathbbR_+^* times mathbbR)$. It is not closed because one can take the family $f(t) = (t,1)$, which belongs to $S$ for $t>0$, and converges to $(0,1) notin S$ as $t$ goes to $0$.
The corresponding family of parameters $lambda (t)$ and $k(t)$ are:
$$lambda(t) = frac1+t^22t, quad k(t) = frac2t1+t^2 (t,1).$$
As $t$ goes to $0$, we have that $k(t)$ converges to $0$. But $lambda (t) k(t)$ does not converge to $0$, because $lambda(t)$ increases fast enough to compensate for the decay of $k(t)$.
hence your mistake is there: the fact that $(k_n)$ converges to $0$ does not imply that $(lambda_n k_n)$ also converges to $0$, because $lambda_n$ has no reason to be bounded.
answered 6 hours ago
D. ThomineD. Thomine
7,7491538
7,7491538
$begingroup$
You are right, it's more clear to me what's happening now. Thank you.
$endgroup$
– mouargmouarg
6 hours ago
add a comment |
$begingroup$
You are right, it's more clear to me what's happening now. Thank you.
$endgroup$
– mouargmouarg
6 hours ago
$begingroup$
You are right, it's more clear to me what's happening now. Thank you.
$endgroup$
– mouargmouarg
6 hours ago
$begingroup$
You are right, it's more clear to me what's happening now. Thank you.
$endgroup$
– mouargmouarg
6 hours ago
add a comment |
$begingroup$
If $ k = 0$ then the sequence $(lambda_n k_n)$ converges to $0 in B$ and we are done.
This is wrong. We know $k_phi(n)to 0$, but $lambda_n$ may be getting large so $lambda_nk_n$ can converge to a nonzero value.
$endgroup$
$begingroup$
Oh, thank you. I should have spotted that... For example by taking : $k_n = frac1n$ and $lambda_n = n$
$endgroup$
– mouargmouarg
6 hours ago
add a comment |
$begingroup$
If $ k = 0$ then the sequence $(lambda_n k_n)$ converges to $0 in B$ and we are done.
This is wrong. We know $k_phi(n)to 0$, but $lambda_n$ may be getting large so $lambda_nk_n$ can converge to a nonzero value.
$endgroup$
$begingroup$
Oh, thank you. I should have spotted that... For example by taking : $k_n = frac1n$ and $lambda_n = n$
$endgroup$
– mouargmouarg
6 hours ago
add a comment |
$begingroup$
If $ k = 0$ then the sequence $(lambda_n k_n)$ converges to $0 in B$ and we are done.
This is wrong. We know $k_phi(n)to 0$, but $lambda_n$ may be getting large so $lambda_nk_n$ can converge to a nonzero value.
$endgroup$
If $ k = 0$ then the sequence $(lambda_n k_n)$ converges to $0 in B$ and we are done.
This is wrong. We know $k_phi(n)to 0$, but $lambda_n$ may be getting large so $lambda_nk_n$ can converge to a nonzero value.
answered 6 hours ago
Eric WofseyEric Wofsey
189k14216347
189k14216347
$begingroup$
Oh, thank you. I should have spotted that... For example by taking : $k_n = frac1n$ and $lambda_n = n$
$endgroup$
– mouargmouarg
6 hours ago
add a comment |
$begingroup$
Oh, thank you. I should have spotted that... For example by taking : $k_n = frac1n$ and $lambda_n = n$
$endgroup$
– mouargmouarg
6 hours ago
$begingroup$
Oh, thank you. I should have spotted that... For example by taking : $k_n = frac1n$ and $lambda_n = n$
$endgroup$
– mouargmouarg
6 hours ago
$begingroup$
Oh, thank you. I should have spotted that... For example by taking : $k_n = frac1n$ and $lambda_n = n$
$endgroup$
– mouargmouarg
6 hours ago
add a comment |
mouargmouarg is a new contributor. Be nice, and check out our Code of Conduct.
mouargmouarg is a new contributor. Be nice, and check out our Code of Conduct.
mouargmouarg is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
I guess you mean $B$ to be the union of the $lambda K$ rather than the set of them?
$endgroup$
– Eric Wofsey
6 hours ago
$begingroup$
@EricWofsey By this notation I mean the set of $lambda x$ for all $x in K$ and $lambda geq 0$, so my notation is wrong ?
$endgroup$
– mouargmouarg
6 hours ago
$begingroup$
Yes, your notation meant that $B$ is the set whose elements are the sets $lambda K$ (so an element of $B$ is a subset of $E$, not a point of $E$).
$endgroup$
– Eric Wofsey
6 hours ago
$begingroup$
@EricWofsey You are right thank you.
$endgroup$
– mouargmouarg
6 hours ago
$begingroup$
Note that you state that $mathbbR^+$ is closed, which isn’t the case
$endgroup$
– Santana Afton
51 mins ago