compactness of a set where am I going wrongIn a metric space, compactness implies completnessClosure and...
How to deal with taxi scam when on vacation?
Welcoming 2019 Pi day: How to draw the letter π?
Can a druid choose the size of its wild shape beast?
What do Xenomorphs eat in the Alien series?
Does someone need to be connected to my network to sniff HTTP requests?
Why doesn't the EU now just force the UK to choose between referendum and no-deal?
Why did it take so long to abandon sail after steamships were demonstrated?
How to use deus ex machina safely?
How do anti-virus programs start at Windows boot?
What did Alexander Pope mean by "Expletives their feeble Aid do join"?
PTIJ: Who should I vote for? (21st Knesset Edition)
A Cautionary Suggestion
Are all passive ability checks floors for active ability checks?
It's a yearly task, alright
Python if-else code style for reduced code for rounding floats
Are there other languages, besides English, where the indefinite (or definite) article varies based on sound?
Time travel from stationary position?
Can I use USB data pins as power source
Why does Bach not break the rules here?
Should we release the security issues we found in our product as CVE or we can just update those on weekly release notes?
Sailing the cryptic seas
Employee lack of ownership
Interplanetary conflict, some disease destroys the ability to understand or appreciate music
Is it true that good novels will automatically sell themselves on Amazon (and so on) and there is no need for one to waste time promoting?
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 $mathbb{R}^n$ 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 $F = {lambda x|lambdain mathbb{R}_+, xin K}$ 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 $B = {lambda x mid lambda in mathbb{R}^+, x in K }$ 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
mouargmouarg is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 3 hours ago
mouargmouarg
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
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 |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
mouargmouarg is a new contributor. Be nice, and check out our Code of Conduct.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3149871%2fcompactness-of-a-set-where-am-i-going-wrong%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
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 |
$begingroup$
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
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.
mouargmouarg is a new contributor. Be nice, and check out our Code of Conduct.
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3149871%2fcompactness-of-a-set-where-am-i-going-wrong%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
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