Mixing
True or false propositions about Compact sets
$begingroup$
Decide whether the following propositions are true or false. If the claim is valid, supply a short proof, and if the claim is false, provide a counterexample.
(a) The arbitrary intersection of compact sets is compact.
(b) The arbitrary union of compact sets is compact.
(c) Let A be arbitrary and let K be compact, then the intersection $Abigcap K$ is compact.
(d) If $F_{1}supseteq F_{2} supseteq F_{3} supseteq F_{4} cdotcdot cdotcdot $is a nested sequence of nonempty closed sets, then the intersection $bigcap_{n=1}^infty F_{n}neq emptyset$
For (a-c) I would like to have my attempt for these solutions checked. Please let me know if these are accurate, not accurate, or accurate but not sufficient. For (d) I need more assistance.
My attempt:
(d) Following the nested interval theorem, this holds. Can someone show me how it does or if I am wrong why it doesn't.
real-analysis proof-verification
asked Nov 9 '15 at 2:41
SelfStudySelfStudy
302216
$endgroup$
add a comment |
$begingroup$
Decide whether the following propositions are true or false. If the claim is valid, supply a short proof, and if the claim is false, provide a counterexample.
(a) The arbitrary intersection of compact sets is compact.
(b) The arbitrary union of compact sets is compact.
(c) Let A be arbitrary and let K be compact, then the intersection $Abigcap K$ is compact.
(d) If $F_{1}supseteq F_{2} supseteq F_{3} supseteq F_{4} cdotcdot cdotcdot $is a nested sequence of nonempty closed sets, then the intersection $bigcap_{n=1}^infty F_{n}neq emptyset$
For (a-c) I would like to have my attempt for these solutions checked. Please let me know if these are accurate, not accurate, or accurate but not sufficient. For (d) I need more assistance.
My attempt:
(d) Following the nested interval theorem, this holds. Can someone show me how it does or if I am wrong why it doesn't.
real-analysis proof-verification
asked Nov 9 '15 at 2:41
SelfStudySelfStudy
302216
$endgroup$
$begingroup$
For C: in $mathbb R$, is $[1,3]cap(2,4)$ compact?
$endgroup$
– Jesse P Francis
Nov 9 '15 at 2:50
1
$begingroup$
Your reasoning in (a) is wrong. In metric spaces, compact sets are bounded and closed, although the reverse is not true in general. The Heine-Borel Theorem applies only when you're working in $mathbb{R}^n$.
$endgroup$
– EA304GT
Nov 9 '15 at 2:54
add a comment |
$begingroup$
Decide whether the following propositions are true or false. If the claim is valid, supply a short proof, and if the claim is false, provide a counterexample.
(a) The arbitrary intersection of compact sets is compact.
(b) The arbitrary union of compact sets is compact.
(c) Let A be arbitrary and let K be compact, then the intersection $Abigcap K$ is compact.
(d) If $F_{1}supseteq F_{2} supseteq F_{3} supseteq F_{4} cdotcdot cdotcdot $is a nested sequence of nonempty closed sets, then the intersection $bigcap_{n=1}^infty F_{n}neq emptyset$
For (a-c) I would like to have my attempt for these solutions checked. Please let me know if these are accurate, not accurate, or accurate but not sufficient. For (d) I need more assistance.
My attempt:
(d) Following the nested interval theorem, this holds. Can someone show me how it does or if I am wrong why it doesn't.
real-analysis proof-verification
asked Nov 9 '15 at 2:41
SelfStudySelfStudy
302216
$endgroup$
Decide whether the following propositions are true or false. If the claim is valid, supply a short proof, and if the claim is false, provide a counterexample.
(a) The arbitrary intersection of compact sets is compact.
(b) The arbitrary union of compact sets is compact.
(c) Let A be arbitrary and let K be compact, then the intersection $Abigcap K$ is compact.
(d) If $F_{1}supseteq F_{2} supseteq F_{3} supseteq F_{4} cdotcdot cdotcdot $is a nested sequence of nonempty closed sets, then the intersection $bigcap_{n=1}^infty F_{n}neq emptyset$
For (a-c) I would like to have my attempt for these solutions checked. Please let me know if these are accurate, not accurate, or accurate but not sufficient. For (d) I need more assistance.
My attempt:
(d) Following the nested interval theorem, this holds. Can someone show me how it does or if I am wrong why it doesn't.
real-analysis proof-verification
real-analysis proof-verification
asked Nov 9 '15 at 2:41
SelfStudySelfStudy
302216
asked Nov 9 '15 at 2:41
SelfStudySelfStudy
302216
edited Nov 9 '15 at 2:47
SelfStudy
asked Nov 9 '15 at 2:41
SelfStudySelfStudy
302216
asked Nov 9 '15 at 2:41
SelfStudySelfStudy
302216
asked Nov 9 '15 at 2:41
SelfStudySelfStudy
302216
302216
$begingroup$
For C: in $mathbb R$, is $[1,3]cap(2,4)$ compact?
$endgroup$
– Jesse P Francis
Nov 9 '15 at 2:50
1
$begingroup$
Your reasoning in (a) is wrong. In metric spaces, compact sets are bounded and closed, although the reverse is not true in general. The Heine-Borel Theorem applies only when you're working in $mathbb{R}^n$.
$endgroup$
– EA304GT
Nov 9 '15 at 2:54
add a comment |
$begingroup$
For C: in $mathbb R$, is $[1,3]cap(2,4)$ compact?
$endgroup$
– Jesse P Francis
Nov 9 '15 at 2:50
1
$begingroup$
Your reasoning in (a) is wrong. In metric spaces, compact sets are bounded and closed, although the reverse is not true in general. The Heine-Borel Theorem applies only when you're working in $mathbb{R}^n$.
$endgroup$
– EA304GT
Nov 9 '15 at 2:54
$begingroup$
For C: in $mathbb R$, is $[1,3]cap(2,4)$ compact?
$endgroup$
– Jesse P Francis
Nov 9 '15 at 2:50
$begingroup$
For C: in $mathbb R$, is $[1,3]cap(2,4)$ compact?
$endgroup$
– Jesse P Francis
Nov 9 '15 at 2:50
1
1
$begingroup$
Your reasoning in (a) is wrong. In metric spaces, compact sets are bounded and closed, although the reverse is not true in general. The Heine-Borel Theorem applies only when you're working in $mathbb{R}^n$.
$endgroup$
– EA304GT
Nov 9 '15 at 2:54
$begingroup$
Your reasoning in (a) is wrong. In metric spaces, compact sets are bounded and closed, although the reverse is not true in general. The Heine-Borel Theorem applies only when you're working in $mathbb{R}^n$.
$endgroup$
– EA304GT
Nov 9 '15 at 2:54
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
For c, you are wrong. Take for example the set $A=(frac{1}{2},1)$ and $K=[0,2]$. Then their intersection is just $A$ which is not compact. As for d), you need a complete metric space, so in the general case it isn't true.
answered Nov 9 '15 at 2:48
EsoogEsoog
35219
$endgroup$
add a comment |
$begingroup$
For $(d)$ In complete metric space, it is known as Cantor's intersection theorem. See this
For $(c)$ take any open set and a compact set and see there intersection. What they say?
answered Nov 9 '15 at 2:46
Kushal BhuyanKushal Bhuyan
4,98421244
$endgroup$
$begingroup$
Oh that's what I was missing. How do (a-c) look?
$endgroup$
– SelfStudy
Nov 9 '15 at 2:54
$begingroup$
$(b)$ looks fine but I have to check $(a)$.
$endgroup$
– Kushal Bhuyan
Nov 9 '15 at 2:55
$begingroup$
As @EA304GT indicated, you can only apply that reasoning for $(a)$ if you are working with $mathbb{R^n}$
$endgroup$
– Kushal Bhuyan
Nov 9 '15 at 3:00
$begingroup$
For arbitrary metric space, Heine Borel theorem: A subset of a metric space is compact if and only if it is complete and totally bounded.
$endgroup$
– Kushal Bhuyan
Nov 9 '15 at 3:03
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
});
}
});
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%2f1520171%2ftrue-or-false-propositions-about-compact-sets%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
$begingroup$
For c, you are wrong. Take for example the set $A=(frac{1}{2},1)$ and $K=[0,2]$. Then their intersection is just $A$ which is not compact. As for d), you need a complete metric space, so in the general case it isn't true.
answered Nov 9 '15 at 2:48
EsoogEsoog
35219
$endgroup$
add a comment |
$begingroup$
For c, you are wrong. Take for example the set $A=(frac{1}{2},1)$ and $K=[0,2]$. Then their intersection is just $A$ which is not compact. As for d), you need a complete metric space, so in the general case it isn't true.
answered Nov 9 '15 at 2:48
EsoogEsoog
35219
$endgroup$
add a comment |
$begingroup$
For c, you are wrong. Take for example the set $A=(frac{1}{2},1)$ and $K=[0,2]$. Then their intersection is just $A$ which is not compact. As for d), you need a complete metric space, so in the general case it isn't true.
answered Nov 9 '15 at 2:48
EsoogEsoog
35219
$endgroup$
For c, you are wrong. Take for example the set $A=(frac{1}{2},1)$ and $K=[0,2]$. Then their intersection is just $A$ which is not compact. As for d), you need a complete metric space, so in the general case it isn't true.
answered Nov 9 '15 at 2:48
EsoogEsoog
35219
answered Nov 9 '15 at 2:48
EsoogEsoog
35219
answered Nov 9 '15 at 2:48
EsoogEsoog
35219
answered Nov 9 '15 at 2:48
EsoogEsoog
35219
35219
add a comment |
add a comment |
$begingroup$
For $(d)$ In complete metric space, it is known as Cantor's intersection theorem. See this
For $(c)$ take any open set and a compact set and see there intersection. What they say?
answered Nov 9 '15 at 2:46
Kushal BhuyanKushal Bhuyan
4,98421244
$endgroup$
$begingroup$
Oh that's what I was missing. How do (a-c) look?
$endgroup$
– SelfStudy
Nov 9 '15 at 2:54
$begingroup$
$(b)$ looks fine but I have to check $(a)$.
$endgroup$
– Kushal Bhuyan
Nov 9 '15 at 2:55
$begingroup$
As @EA304GT indicated, you can only apply that reasoning for $(a)$ if you are working with $mathbb{R^n}$
$endgroup$
– Kushal Bhuyan
Nov 9 '15 at 3:00
$begingroup$
For arbitrary metric space, Heine Borel theorem: A subset of a metric space is compact if and only if it is complete and totally bounded.
$endgroup$
– Kushal Bhuyan
Nov 9 '15 at 3:03
add a comment |
$begingroup$
For $(d)$ In complete metric space, it is known as Cantor's intersection theorem. See this
For $(c)$ take any open set and a compact set and see there intersection. What they say?
answered Nov 9 '15 at 2:46
Kushal BhuyanKushal Bhuyan
4,98421244
$endgroup$
$begingroup$
Oh that's what I was missing. How do (a-c) look?
$endgroup$
– SelfStudy
Nov 9 '15 at 2:54
$begingroup$
$(b)$ looks fine but I have to check $(a)$.
$endgroup$
– Kushal Bhuyan
Nov 9 '15 at 2:55
$begingroup$
As @EA304GT indicated, you can only apply that reasoning for $(a)$ if you are working with $mathbb{R^n}$
$endgroup$
– Kushal Bhuyan
Nov 9 '15 at 3:00
$begingroup$
For arbitrary metric space, Heine Borel theorem: A subset of a metric space is compact if and only if it is complete and totally bounded.
$endgroup$
– Kushal Bhuyan
Nov 9 '15 at 3:03
add a comment |
$begingroup$
For $(d)$ In complete metric space, it is known as Cantor's intersection theorem. See this
For $(c)$ take any open set and a compact set and see there intersection. What they say?
answered Nov 9 '15 at 2:46
Kushal BhuyanKushal Bhuyan
4,98421244
$endgroup$
For $(d)$ In complete metric space, it is known as Cantor's intersection theorem. See this
For $(c)$ take any open set and a compact set and see there intersection. What they say?
answered Nov 9 '15 at 2:46
Kushal BhuyanKushal Bhuyan
4,98421244
edited Nov 9 '15 at 2:52
answered Nov 9 '15 at 2:46
Kushal BhuyanKushal Bhuyan
4,98421244
answered Nov 9 '15 at 2:46
Kushal BhuyanKushal Bhuyan
4,98421244
answered Nov 9 '15 at 2:46
Kushal BhuyanKushal Bhuyan
4,98421244
4,98421244
$begingroup$
Oh that's what I was missing. How do (a-c) look?
$endgroup$
– SelfStudy
Nov 9 '15 at 2:54
$begingroup$
$(b)$ looks fine but I have to check $(a)$.
$endgroup$
– Kushal Bhuyan
Nov 9 '15 at 2:55
$begingroup$
As @EA304GT indicated, you can only apply that reasoning for $(a)$ if you are working with $mathbb{R^n}$
$endgroup$
– Kushal Bhuyan
Nov 9 '15 at 3:00
$begingroup$
For arbitrary metric space, Heine Borel theorem: A subset of a metric space is compact if and only if it is complete and totally bounded.
$endgroup$
– Kushal Bhuyan
Nov 9 '15 at 3:03
add a comment |
$begingroup$
Oh that's what I was missing. How do (a-c) look?
$endgroup$
– SelfStudy
Nov 9 '15 at 2:54
$begingroup$
$(b)$ looks fine but I have to check $(a)$.
$endgroup$
– Kushal Bhuyan
Nov 9 '15 at 2:55
$begingroup$
As @EA304GT indicated, you can only apply that reasoning for $(a)$ if you are working with $mathbb{R^n}$
$endgroup$
– Kushal Bhuyan
Nov 9 '15 at 3:00
$begingroup$
For arbitrary metric space, Heine Borel theorem: A subset of a metric space is compact if and only if it is complete and totally bounded.
$endgroup$
– Kushal Bhuyan
Nov 9 '15 at 3:03
$begingroup$
Oh that's what I was missing. How do (a-c) look?
$endgroup$
– SelfStudy
Nov 9 '15 at 2:54
$begingroup$
Oh that's what I was missing. How do (a-c) look?
$endgroup$
– SelfStudy
Nov 9 '15 at 2:54
$begingroup$
$(b)$ looks fine but I have to check $(a)$.
$endgroup$
– Kushal Bhuyan
Nov 9 '15 at 2:55
$begingroup$
$(b)$ looks fine but I have to check $(a)$.
$endgroup$
– Kushal Bhuyan
Nov 9 '15 at 2:55
$begingroup$
As @EA304GT indicated, you can only apply that reasoning for $(a)$ if you are working with $mathbb{R^n}$
$endgroup$
– Kushal Bhuyan
Nov 9 '15 at 3:00
$begingroup$
As @EA304GT indicated, you can only apply that reasoning for $(a)$ if you are working with $mathbb{R^n}$
$endgroup$
– Kushal Bhuyan
Nov 9 '15 at 3:00
$begingroup$
For arbitrary metric space, Heine Borel theorem: A subset of a metric space is compact if and only if it is complete and totally bounded.
$endgroup$
– Kushal Bhuyan
Nov 9 '15 at 3:03
$begingroup$
For arbitrary metric space, Heine Borel theorem: A subset of a metric space is compact if and only if it is complete and totally bounded.
$endgroup$
– Kushal Bhuyan
Nov 9 '15 at 3:03
add a comment |
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%2f1520171%2ftrue-or-false-propositions-about-compact-sets%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
$begingroup$
For C: in $mathbb R$, is $[1,3]cap(2,4)$ compact?
$endgroup$
– Jesse P Francis
Nov 9 '15 at 2:50
1
$begingroup$
Your reasoning in (a) is wrong. In metric spaces, compact sets are bounded and closed, although the reverse is not true in general. The Heine-Borel Theorem applies only when you're working in $mathbb{R}^n$.
$endgroup$
– EA304GT
Nov 9 '15 at 2:54