Mathematics Stack Exchange
Mathematics Stack Exchange

Questions tagged [compactness]

The compactness tag is for questions about compactness and its many variants (e.g. sequential compactness, countable compactness) as well locally compact spaces; compactifications (e.g. one-point, Stone-Čech) and other topics closely related to compactness. This includes logical compactness.

Compactness is a topological property. We say that a topological space $X$ is compact if whenever we cover $X$ by a collection of open sets we can find a finite number of open sets from the collection which cover $X$. For example, $[0,1]$ is a compact subspace of $\mathbb{R}$, but $(0,1)$ and $\mathbb{R}$ are not.

We say that a space $X$ is sequentially compact if every sequence has a convergent subsequence. These properties are equivalent for metric space, although neither implies the other in general.

This tag may also be used for questions about logical compactness, such as the compactness theorem.

More information can be found on Wikipedia.

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Why is compactness so important?

I've read many times that 'compactness' is such an extremely important and useful concept, though it's still not very apparent why. The only theorems I've seen concerning it are the Heine-Borel theorem, and a proof continuous functions on R from…
FireGarden
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What should be the intuition when working with compactness?

I have a question that may be regarded by many as duplicate since there's a similar one at MathOverflow. In $\mathbb{R}^n$ the compact sets are those that are closed and bounded, however the guy who answered this question and had his answer…
Gold
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$A$ and $B$ disjoint, $A$ compact, and $B$ closed implies there is positive distance between both sets.

Claim: Let $X$ be a metric space. If $A,B\subset X$ are disjoint, $A$ is compact, and $B$ is closed, then there is $\delta>0$ so that $ |\alpha-\beta|\geq\delta\;\;\;\forall\alpha\in A,\beta\in B$. Proof. Assume the contrary. Let $\alpha_n\in…
Benji
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Projection map being a closed map

Let $\pi: X \times Y \to X$ be the projection map where $Y$ is compact. Prove that $\pi$ is a closed map. First I would like to see a proof of this claim. I want to know that here why compactness is necessary or do we have any other weaker…
M.Subramani
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What's going on with "compact implies sequentially compact"?

I've seen both counterexamples and proofs to "compact implies sequentially compact", and I'm not sure what's going on. Apparently there are compact spaces which are not sequentially compact; quick googling and wikipedia checks will turn up examples…
matt
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Why is compactness in logic called compactness?

In logic, a semantics is said to be compact iff if every finite subset of a set of sentences has a model, then so to does the entire set. Most logic texts either don't explain the terminology, or allude to the topological property of compactness. I…
vanden
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Compact sets are closed?

I feel really ignorant in asking this question but I am really just don't understand how a compact set can be considered closed. By definition of a compact set it means that given an open cover we can find a finite subcover the covers the…
InsigMath
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How to prove every closed interval in R is compact?

Let $[a,b]\subseteq \mathbb R$. As we know, it is compact. This is a very important result. However, the proof for the result may be not familiar to us. Here I want to collect the ways to prove $[a,b]$ is compact. Thanks for your help and any link.
Paul
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Intersection of finite number of compact sets is compact?

Is the the intersection of a finite number of compact sets is compact? If not please give a counter example to demonstrate this is not true. I said that this is true because the intersection of finite number of compact sets are closed. Which…
John Buchta
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How to prove that a compact set in a Hausdorff topological space is closed?

How to prove that a compact set $K$ in a Hausdorff topological space $\mathbb{X}$ is closed? I seek a proof that is as self contained as possible. Thank you.
Elias Costa
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If every real-valued continuous function is bounded on $X$ (metric space), then $X$ is compact.

Let $X$ be a metric space. Prove that if every continuous function $f: X \rightarrow \mathbb{R}$ is bounded, then $X$ is compact. This has been asked before, but all the answers I have seen prove the contrapositive. Realistically, this may be the…
combinator
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Difference between closed, bounded and compact sets

In real analysis, there is a theorem that a bounded sequence has a convergent subsequence. Also, the limit lies in the same set as the elements of the sequence, if the set is closed. Then when metric spaces are introduced, there is a similar…
user125395
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Prove if $X$ is a compact metric space, then $X$ is separable.

Let $X$ be a metric space. Prove if $X$ is compact, then $X$ is separable. X separable $\iff X$ contains a countable dense subset. $E \subset X $ dense in $X \iff \overline{E} = X$. $X$ compact $\iff$ every open cover of $X$ admits a finite…
user181728
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To show that the set point distant by 1 of a compact set has Lebesgue measure $0$

Could any one tell me how to solve this one? Let $K$ be a compact subset of $\mathbb{R}^n$, and $$A:=\{x\in\mathbb{R}^n:d(x,K)=1\}.$$ Show that $A$ has Lebesgue measure $0$. Thank you!
Balbichi
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Totally bounded, complete $\implies$ compact

Show that a totally bounded complete metric space $X$ is compact. I can use the fact that sequentially compact $\Leftrightarrow$ compact. Attempt: Complete $\implies$ every Cauchy sequence converges. Totally bounded $\implies$…
Emir
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