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compactness of a set where am I going wrong
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$begingroup$
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 $B = {lambda x mid lambda in mathbb{R}^+, x in K }$ is closed (where $mathbb{R}^+$ 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 $phi : mathbb{N} to mathbb{N}$ 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 $mu in mathbb{R}^*$ such that $k = mu x$. From here we can deduce that the sequence $(lambda_n)$ necessarily converges to $frac{1}{mu}$. Yet since $mathbb{R}^+$ is closed the sequence $(lambda_n)$ converges to a positive real number, so $frac{1}{mu} geq 0$ so $mu geq 0$.
So the sequence $(lambda_n k_n)$ converges to the vector $frac{1}{mu} k in B$ since $k in K$ and $frac{1}{mu} geq 0$. Hence $B$ is closed.
So where am I going wrong here ?
Thank you !
real-analysis general-topology proof-verification compactness
asked 3 hours ago
mouargmouargmouargmouarg
433
New contributor
mouargmouarg is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
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 $B = {lambda x mid lambda in mathbb{R}^+, x in K }$ is closed (where $mathbb{R}^+$ 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 $phi : mathbb{N} to mathbb{N}$ 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 $mu in mathbb{R}^*$ such that $k = mu x$. From here we can deduce that the sequence $(lambda_n)$ necessarily converges to $frac{1}{mu}$. Yet since $mathbb{R}^+$ is closed the sequence $(lambda_n)$ converges to a positive real number, so $frac{1}{mu} geq 0$ so $mu geq 0$.
So the sequence $(lambda_n k_n)$ converges to the vector $frac{1}{mu} k in B$ since $k in K$ and $frac{1}{mu} geq 0$. Hence $B$ is closed.
So where am I going wrong here ?
Thank you !
real-analysis general-topology proof-verification compactness
asked 3 hours ago
mouargmouargmouargmouarg
433
New contributor
mouargmouarg is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
I guess you mean $B$ to be the union of the $lambda K$ rather than the set of them?
$endgroup$
– Eric Wofsey
3 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
3 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
3 hours ago
$begingroup$
@EricWofsey You are right thank you.
$endgroup$
– mouargmouarg
3 hours 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 $B = {lambda x mid lambda in mathbb{R}^+, x in K }$ is closed (where $mathbb{R}^+$ 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 $phi : mathbb{N} to mathbb{N}$ 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 $mu in mathbb{R}^*$ such that $k = mu x$. From here we can deduce that the sequence $(lambda_n)$ necessarily converges to $frac{1}{mu}$. Yet since $mathbb{R}^+$ is closed the sequence $(lambda_n)$ converges to a positive real number, so $frac{1}{mu} geq 0$ so $mu geq 0$.
So the sequence $(lambda_n k_n)$ converges to the vector $frac{1}{mu} k in B$ since $k in K$ and $frac{1}{mu} geq 0$. Hence $B$ is closed.
So where am I going wrong here ?
Thank you !
real-analysis general-topology proof-verification compactness
asked 3 hours ago
mouargmouargmouargmouarg
433
New contributor
mouargmouarg is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
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 $B = {lambda x mid lambda in mathbb{R}^+, x in K }$ is closed (where $mathbb{R}^+$ 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 $phi : mathbb{N} to mathbb{N}$ 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 $mu in mathbb{R}^*$ such that $k = mu x$. From here we can deduce that the sequence $(lambda_n)$ necessarily converges to $frac{1}{mu}$. Yet since $mathbb{R}^+$ is closed the sequence $(lambda_n)$ converges to a positive real number, so $frac{1}{mu} geq 0$ so $mu geq 0$.
So the sequence $(lambda_n k_n)$ converges to the vector $frac{1}{mu} k in B$ since $k in K$ and $frac{1}{mu} geq 0$. 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
asked 3 hours ago
mouargmouargmouargmouarg
433
New contributor
mouargmouarg is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 3 hours ago
mouargmouargmouargmouarg
433
New contributor
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edited 3 hours ago
mouargmouarg
asked 3 hours ago
mouargmouargmouargmouarg
433
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Check out our Code of Conduct.
asked 3 hours ago
mouargmouargmouargmouarg
433
asked 3 hours ago
mouargmouargmouargmouarg
433
433
New contributor
mouargmouarg is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
mouargmouarg is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
mouargmouarg is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
I guess you mean $B$ to be the union of the $lambda K$ rather than the set of them?
$endgroup$
– Eric Wofsey
3 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
3 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
3 hours ago
$begingroup$
@EricWofsey You are right thank you.
$endgroup$
– mouargmouarg
3 hours 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
3 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
3 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
3 hours ago
$begingroup$
@EricWofsey You are right thank you.
$endgroup$
– mouargmouarg
3 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
3 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
3 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
3 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
3 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
3 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
3 hours ago
$begingroup$
@EricWofsey You are right thank you.
$endgroup$
– mouargmouarg
3 hours ago
$begingroup$
@EricWofsey You are right thank you.
$endgroup$
– mouargmouarg
3 hours 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 $S := bigcup_{lambda geq 0} lambda K = {(0,0)} cup (mathbb{R}_+^* times mathbb{R})$. 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:
$$lambda(t) = frac{1+t^2}{2t}, quad k(t) = frac{2t}{1+t^2} (t,1).$$
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 3 hours ago
D. ThomineD. Thomine
7,7291538
$endgroup$
$begingroup$
You are right, it's more clear to me what's happening now. Thank you.
$endgroup$
– mouargmouarg
3 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.
answered 3 hours ago
Eric WofseyEric Wofsey
189k14216347
$endgroup$
$begingroup$
Oh, thank you. I should have spotted that... For example by taking : $k_n = frac{1}{n}$ and $lambda_n = n$
$endgroup$
– mouargmouarg
3 hours ago
add a comment |
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2 Answers
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2 Answers
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oldest
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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 $S := bigcup_{lambda geq 0} lambda K = {(0,0)} cup (mathbb{R}_+^* times mathbb{R})$. 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:
$$lambda(t) = frac{1+t^2}{2t}, quad k(t) = frac{2t}{1+t^2} (t,1).$$
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 3 hours ago
D. ThomineD. Thomine
7,7291538
$endgroup$
$begingroup$
You are right, it's more clear to me what's happening now. Thank you.
$endgroup$
– mouargmouarg
3 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 $S := bigcup_{lambda geq 0} lambda K = {(0,0)} cup (mathbb{R}_+^* times mathbb{R})$. 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:
$$lambda(t) = frac{1+t^2}{2t}, quad k(t) = frac{2t}{1+t^2} (t,1).$$
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 3 hours ago
D. ThomineD. Thomine
7,7291538
$endgroup$
$begingroup$
You are right, it's more clear to me what's happening now. Thank you.
$endgroup$
– mouargmouarg
3 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 $S := bigcup_{lambda geq 0} lambda K = {(0,0)} cup (mathbb{R}_+^* times mathbb{R})$. 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:
$$lambda(t) = frac{1+t^2}{2t}, quad k(t) = frac{2t}{1+t^2} (t,1).$$
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 3 hours ago
D. ThomineD. Thomine
7,7291538
$endgroup$
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 $S := bigcup_{lambda geq 0} lambda K = {(0,0)} cup (mathbb{R}_+^* times mathbb{R})$. 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:
$$lambda(t) = frac{1+t^2}{2t}, quad k(t) = frac{2t}{1+t^2} (t,1).$$
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 3 hours ago
D. ThomineD. Thomine
7,7291538
answered 3 hours ago
D. ThomineD. Thomine
7,7291538
answered 3 hours ago
D. ThomineD. Thomine
7,7291538
answered 3 hours ago
D. ThomineD. Thomine
7,7291538
7,7291538
$begingroup$
You are right, it's more clear to me what's happening now. Thank you.
$endgroup$
– mouargmouarg
3 hours ago
add a comment |
$begingroup$
You are right, it's more clear to me what's happening now. Thank you.
$endgroup$
– mouargmouarg
3 hours ago
$begingroup$
You are right, it's more clear to me what's happening now. Thank you.
$endgroup$
– mouargmouarg
3 hours ago
$begingroup$
You are right, it's more clear to me what's happening now. Thank you.
$endgroup$
– mouargmouarg
3 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.
answered 3 hours ago
Eric WofseyEric Wofsey
189k14216347
$endgroup$
$begingroup$
Oh, thank you. I should have spotted that... For example by taking : $k_n = frac{1}{n}$ and $lambda_n = n$
$endgroup$
– mouargmouarg
3 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.
answered 3 hours ago
Eric WofseyEric Wofsey
189k14216347
$endgroup$
$begingroup$
Oh, thank you. I should have spotted that... For example by taking : $k_n = frac{1}{n}$ and $lambda_n = n$
$endgroup$
– mouargmouarg
3 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.
answered 3 hours ago
Eric WofseyEric Wofsey
189k14216347
$endgroup$
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 3 hours ago
Eric WofseyEric Wofsey
189k14216347
answered 3 hours ago
Eric WofseyEric Wofsey
189k14216347
answered 3 hours ago
Eric WofseyEric Wofsey
189k14216347
answered 3 hours ago
Eric WofseyEric Wofsey
189k14216347
189k14216347
$begingroup$
Oh, thank you. I should have spotted that... For example by taking : $k_n = frac{1}{n}$ and $lambda_n = n$
$endgroup$
– mouargmouarg
3 hours ago
add a comment |
$begingroup$
Oh, thank you. I should have spotted that... For example by taking : $k_n = frac{1}{n}$ and $lambda_n = n$
$endgroup$
– mouargmouarg
3 hours ago
$begingroup$
Oh, thank you. I should have spotted that... For example by taking : $k_n = frac{1}{n}$ and $lambda_n = n$
$endgroup$
– mouargmouarg
3 hours ago
$begingroup$
Oh, thank you. I should have spotted that... For example by taking : $k_n = frac{1}{n}$ and $lambda_n = n$
$endgroup$
– mouargmouarg
3 hours ago
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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
3 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
3 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
3 hours ago
$begingroup$
@EricWofsey You are right thank you.
$endgroup$
– mouargmouarg
3 hours ago