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.

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If $\mathcal{F}$ is an equicontinuous family of function from $\mathbb{X}$, compact, on $\mathbb{Y}$ then $\mathcal{F}$ is uniformly equicontinuous

Let ($\mathbb{X}$,d) and ($\mathbb{Y}$,d') metric spaces. $\mathbb{X}$ is compact. Prove that: if $\mathcal{F}$ is an equicontinuous family of function from $\mathbb{X}$ to $\mathbb{Y}$ then ...
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27 views

compactness of closed interval

Using the topological definition of compact, namely that every open cover admits a finite open subcover, I was hoping someone can provide me with the finite sub cover of ...
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1answer
48 views

Any open cover of $S^1$ is an open cover of the annulas

The question goes like this : If $\{U_i:i\in I\}$ is an open cover of the unit circle in $\mathbb R^2$ then show that it is an open cover of an annulus $1-\delta\lt ...
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41 views

The countable dense subset of every compact metric space

Show that any compact metric space has a countable dense subset. I am having problem with finishing the proof after a few steps. This is how I am going : So, let $X$ be the compact ...
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1answer
28 views

If $A$ is not totally bounded then $A$ has an infinite subset $B$ homeomorphic to a discrete space

If $A$ is not totally bounded then $A$ has an infinite subset $B$ homeomorphic to a discrete space. My approach since $A$ is not totally bounded we can find $\epsilon>0$ and a sequence $x_n$ ...
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37 views

Showing that a subset of $\ell^1$ is totally bounded

Let $A=\{ f \in \ell^1\, :\,\text{ for each natural number}\, n\,\text{we have}\, |f(n)|<1/2^n \}$. Show that it is totally bounded in $\ell^1$ and find the interior of $A$ in $\ell^1$ the ...
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32 views

Fixed “$\mathcal Set$” for continuous maps on compact spaces

Let $$f:X\rightarrow X$$ be a continuous map on the compact metric space $X.$ Show that there is a subset $A\subset X$ such that $f(A)=A$. Now the given hint is that to ...
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43 views

A step in proving quasicomponent and component in compact Hausdorff space are the same

I'm trying to prove this lemma from exercise of "General Toplology" book(James Munkres): Let X be a compact Hausdorff space. Then $x, y$ belong to the same quasicomponent of $X$ iff they belong to ...
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Thms on Dynamical Systems: Cont. functions on Compact Spaces (sources)

I've recently started taking a discrete-time dynamical systems perspective on a topic. I've been able to introduce a reasonable metric on a set, obtaining a compact space. Under that metric, a nice ...
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24 views

first countable, anti compact which is not Hausdorff

A space $X$ is called anti-compact if any compact subset of $X$ is finite. In this answer, there are three anti compact Haudorff which is not discrete and in this answer, Brian M. Scott has gave anti ...
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1answer
59 views

Is the set $C = \{(x, y, z): x^2 + y^2 + z^2 = \pi\}$ compact?

I am working on a problem in which I must determine if the extrema (I found these in the initial part of the problem) of a function $f: \mathbb{R}^3 \to \mathbb{R}$, constrained to the set $$C = ...
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42 views

Question about maximal collection with finite intersection property

I'm reading "General Topology" of James Munkres book, chapter 5 about Tychonoff theorem and got stuck on this problem: Let $X$ be a space. Let $\mathscr{D}$ be a collection of subsets of $X$ ...
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1answer
22 views

Inverse that isn't Sequentially Compact

I am looking for a function $f: R\rightarrow R$ and a sequentially compact $K\subset R$ such that the inverse $f^{-1}(K)$ is not sequentially compact. I decided to choose $f(x) = sin(x)$, but I'm ...
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21 views

Finding a continuous function and a sequentially compact

Is there such a continuous function $f: R\rightarrow R $ and a sequentially compact $K \subset R$ such that the inverse $f^{-1}(K)$ is not sequentially compact? Could someone provide me with some ...
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1answer
44 views

A nowhere continuous function that maps compact sets to compact sets

Construct an example of a function $f:\mathbb{R}\to \mathbb{R}$ that is not continuous at any point, but satisfies the property "$f(K)$ is compact, when $K$ is compact" however $f(\mathbb{R})$ is ...
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20 views

Special linear group is for $n\geq 1$ closed, but for $n=1$ also compact

Let be $$SL_n(\mathbb{R}):=\{A\in \mathbb{R}^{n\times n}|\det(A)=1\},\quad n\geq 1$$ a special linear group. Prove that $SL_n(\mathbb{R})$ is for $n\geq 1$ closed in $\mathbb{R}^{n\times ...
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Is this definition of normal family equivalent to the following?

Definition Let $G$ be open in $\mathbb{C}$ and $Y$ be a complete metric space and $\mathscr{F}\subset C(G,Y)$. Then, $\mathscr{F}$ is normal if every sequence of functions in ...
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29 views

Basic understanding of compact sets in regards to covers

So before my book even goes into the theorem that says "A set $E \subset R $ is compact iff E is closed and bounded", they go into examples of an open set, but I'm having trouble understanding their ...
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29 views

Prove if $C$ and $D$ are non empty disjoint closed subsets of $M$ such that $C$ or $D$ is compact, then $d(C,D)$ > 0

Prove if $C$ and $D$ are non empty disjoint closed subsets of $M$ such that $C$ or $D$ is compact, then $d(C,D)$ > 0 In $R$, two closed subsets are compact and their distance is definitely greater ...
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1answer
53 views

Proving that a one-to-one continuous function on a compact subset has a continuous inverse

This is a curious problem I found in the "challenge" section of the text I'm learning real analysis from. Suppose $A$ is a compact subset of $\mathbb{R}^{n}$ and that $f$ is a continuous function ...
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1answer
20 views

finding noninterval sets that are either closed and unbounded or bounded and open

I need to find an example of a closed set $E$ that is not an interval and not compact and a bounded set $F$ that is not an interval and not compact. I'm really confused how to find this without using ...
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1answer
34 views

uncountable subcovers of a compact set

Example of a compact set $K$ and a cover of it by closed sets such that the cover contains no countable subcover of $K$. So my $K$ would be $[0,1]$ and my cover would be ...
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compactness theorem - first order logic

A basic question that keeps bugging me. Using compactness theorem, we can show that there exists a model of $Th(\mathbb N)$ [with the signature {<}] with an infinite element. But in $\mathbb N$ it ...
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28 views

Show that C^α([0, 1]) is of first category in C([0, 1]).

Recall that for any $0 < α < 1,$ the space $C^\alpha ([0, 1])$ is the set of continuous functions on $[0, 1]$ with norm of f = sup |f| + $ sup \ x\ne y$ |f(x) − f(y)|/|x − y|^α< ∞, equipped ...
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56 views

Is the long line completely uniformizable?

The long line $L$ is uniformizable; in fact, as $L$ is a locally compact Hausdorff space we can explicitly write down a uniformity for it: If $\hat{L}$ is the one-point compactification of $L$, then ...
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Every sequence of real numbers can be covered by a sequence of open intervals of arbitrarily small length.

How do I see that every sequence of real numbers can be covered by a sequence of open intervals of arbitrarily small length, namely that if $x_1, x_2, x_3, \dots$ is a sequence of real numbers and ...
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18 views

If k is a compact subset of R and g(k) is a continuous function from R to R, Show that g(k) is also a compact set.

If k is a compact subset of R and g(k) is a continuous function from R to R, Show that g(k) is also a compact set. Attempt Let $U_{\alpha}$ be any open cover of g(k), then $g^{-1}(U_{\alpha})$ is an ...
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22 views

Two sets S and T are disjoint and compact in a normed vector space. Define $f(S,T)=\inf\{||s-t||:s \in S, t \in T\}$. [duplicate]

Two sets S and T are disjoint and compact in a normed vector space. Define $f(S,T)=\inf\{||s-t||:s \in S, t \in T\}$. Are there elements $s \in S$ and $t \in T$ s.t. $f(S,T)=||s-t||$?
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Let $A$, $B$ be compact subsets of $\Bbb R$ and define $C = A + B =\{\, a+b\mid a \in A , b \in B\,\}$.

Let $A$, $B$ be compact subsets of $\Bbb{R}$ and define $C = A + B =\{\, a+b\mid a \in A , b \in B\,\}$. Prove that $C$ is compact. Use induction to show if $A_1, A_2,\ldots A_n$ are compact subsets ...
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26 views

Closed and bounded but not compact in $L^p(\mathbb{R}^n)$

I am reading the paper "The Kolmogorov-Riesz Compactness Theorem", which gives a characterisation for totally bounded subsets of $L^P(\mathbb{R}^n)$ for $1 \le p <\infty$. A subset $\mathcal{F} ...
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50 views

Are compact subsets of metric spaces closed and bounded?

Rubin's Theorem 2.34 says that "Compact subsets of metric spaces are closed." But I'm wondering should it also be true that, compact subsets of metric spaces are closed and bounded at the same time? ...
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$K$ is compact to $X$ if and only $K$ is compact to $Y$ even if $X$ and $Y$ is not subset of one another?

Is the following statement true: $K \subset X$ and $K \subset Y$ such that $X \cap Y \ne \varnothing$ and $X \not\subset Y, Y \not\subset X$, then $K$ is compact relative to $X$ if and only if $K$ is ...
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Show that the action of $G$ over $X\cong G/K$ is proper for $K$ compact

During a proof on Brown's book (details below) I encountered the following claim Let $G$ be a topological group that acts transitively on a manifold $X$, and let $K$ be the isotropy group. Assume ...
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Using the definition of compactness, show that $\Omega = [1, 3)$ is not compact

I want to use the definition of a compact set to determine whether the following set $\Omega$ is compact (Clearly, $\Omega$ is not compact due to the Heine-Borel theorem). Let $\Omega = [1, 3)$ and ...
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1answer
64 views

Unit interval in $\mathbb{Q}$ is not totally bounded.

I was reading an analysis textbook, and I came across with a theorem says that a set is compact iff it's closed and totally bounded. But if we consider the unit interval in the metric space ...
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1answer
19 views

Proving a sequence in a compact set contains a point that converges to zero

Let $E$ be a compact set and let $f$ be continuous on $E$. Suppose there is a sequence {$x_n$}$_{n=1 \to \infty}$ of such points of E such that $\lvert f(x_n) \rvert < \frac{1}{n}$ for each $n$. ...
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52 views

Proving that a compact set which has a limit point at each point in the set is bounded

Suppose $f:K \to (-\infty, \infty), K $ is compact, and $f$ has a finite limit at each point of $K$, but may not be continuous on $K$. Show that f is bounded. Then what if we don't know if $f$ is ...
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32 views

Proving the compactness of a set

Suppose $f:D \to R$ is continuous with D compact. Prove that {$x:0 \le f(x) \le 1$} is compact Since D is compact and continuous, we know it is uniformly continuous. Let $E=$ {$x:0 \le f(x) \le 1$}, ...
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Where can I find the paper “Un théorème de compacité” by J. P. Aubin?

Is there any source (online or book) for the paper below? J. P. Aubin, Un théorème de compacité, C.R. Acad. Sc. Paris, 256 (1963), pp. 5042–5044. It seems this paper is the origin of the ...
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103 views

How to fix a proof of Dini's Lemma

I am aware of 2 proofs of Dini. One is by contradiction and Bolzano-Weierstrass, and one is by an open covers definition. I decided to try to make a 'direct' proof without using 'every open cover has ...
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43 views

Prove that a continuous function on $\mathbb{R}$ which is periodic attains its extreme values.

Prove that a continuous function on $\mathbb{R}$ which is periodic attains its extreme values. I don't know how to start this proof, but I know I have to use the extreme value theorem, continuity ...
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1answer
34 views

Show there exists an $f$ s.t. $f(x)>0$ for $x \in C$ where $C$ is compact

I thought to use the same functions $h$ as we used in (c) but the problem is the $h$ is positive on $(a^1-\epsilon,a^1+\epsilon) \times ... \times (a^n - \epsilon, a^n + \epsilon)$ which is not ...
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1answer
32 views

If $A$ is closed and $B$ is compact in $\mathbb R^n$ then $A+B=\{a+b : a \in A \text{ and } b \in B\}$ is closed.

If $A$ is closed and $B$ is compact in $\mathbb R^n$ then $A+B=\{a+b : a \in A \text{ and } b \in B\}$ is closed. (In other words, the vector/Minkowski sum of a closed set and a compact set is ...
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38 views

Locally Compact Metric Space Properties

(i) Let $X$ be locally compact, and $K$ compact. Show that, given $\delta > 0$, there exists finitely many compact closed balls of radius at most $\delta$ that cover $K$. Proof: Let $\delta ...
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Must the index for the sets in a finite open subcover $\mathscr{G}$ be arbitrary or can it be a specific number?

I have a question regarding a proof in my text proving that the set $S = (1,2)$ is not compact without using the Heine-Borel theorem. Note: This is a real analysis related question. Here is the ...
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1answer
49 views

Antiderivative as an integral operator from $L^2(0,2\pi)$ to $L^2(0,2\pi)$

I am starting to study Functional Analysis on Hilbert Spaces and I am studying the following operator: $$T:L^2(0,2\pi) \rightarrow L^2(0,2\pi) $$ where $$Tf:(0,2\pi) \rightarrow \mathbb{R} \\ ...
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23 views

A little question about the proof of *compactness* implying *limit point compactness* .

$X$ is an arbitrary topological space which is compact . We have to prove that $X$ is limit point compact . So , $A$ be any infinite set if $X$ . If we suppose $A$ "has no limit ...
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1answer
41 views

Countably infinite union of compact sets is compact

If $A_n$ are compact in $\mathbb R$ for $n\in\mathbb N$, then $\bigcup_{n=1}^\infty A_n$ is compact. I seem to be able to prove the statement by proving the base case that the union of 2 compact sets ...
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1answer
50 views

Compactness of the closed unit ball of ${\rm Lip}_0(X)$ in the topology of pointwise convergence implies that ${\rm Lip}_0(X)$ is a dual space

It is proven that a closed unit ball in the set of real-valued Lipschitz functions ${\rm Lip}_0(X)$ defined on a Banach space $X$ is compact for the topology of pointwise convergence. However, I ...
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24 views

Determine why these sets are necessarily open, closed or compact where $f$ is a continuous function

Determine why these sets are necessarily open, closed or compact where $f$ is a continuous function $f : \mathbb{R} \to \mathbb{R}$ a) $S=$ { $x \in \mathbb{R} : f(x) = -3$ } {$-3$} is closed thus ...