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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Compactness of Propositional Logic

I little confused on compactness of propositional logic. So propositional logic has the property of being compact, that is to say, given a set of formulas $\mathcal F$, then $\mathcal F$ is ...
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1answer
18 views

Continous function on compact interval - bounded

Let $K$ be a compact interval in $\mathbb{R}$. Then every continous function $\phi :K\rightarrow \mathbb{R}^d$ is automatically bounded. Is this a consequence of; the image of a compact is compact ? ...
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43 views

If an upper semicontinuous multivalued map is compact on a set, is it compact on the boundary as well?

I have stumbled upon the following problem during my research: Let $X$ and $Y$ be Banach spaces, $K\subset X$ nonempty, $F:\overline{K}\rightarrow 2^{Y}$ an upper semicontinuous multivalued map with ...
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1answer
36 views

Compact and connected

Let $\mathbb{J} :=\{1/n: 0< n\in \mathbb{Z}\}$ Let $T_{ir}$ be topology of $\mathbb{R}$ generated by $$\{(a,b)\subset \mathbb{R}:a<b\}\cup\{(a,b) \setminus \mathbb{J}\subset ...
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32 views

Topology - Compactness of $\mathbb{Z}\times\{0,1\}$

A question from my h.w.: Is the topological space $\mathbb{Z}\times\{0,1\}$ (where $\mathbb{Z}$ has the discrete topology and $\{0,1\}$ the trivial one) compact? sequentially compact? ...
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1answer
17 views

Help undestanding compactness with convergent subsequences

One way to define compactness in metric spaces is to note that in compact metric space each sequence has a convergent subsequence. Understanding compactness is difficult for me from this ...
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0answers
34 views

A bijective function $f$ between two compact Hausdorff spaces is continuous if $f$ preserves compact sets [duplicate]

I am trying to prove that if $f: X \longrightarrow Y$ is a bijection between two compact Hausdorff spaces such that $f[W]$ is compact in $Y$ for all compact $W$ in $X$, then $f$ is continuous. Here ...
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1answer
78 views

Example of closed and bounded set that is not compact

Consider the metric space $Q$ of rational numbers with the Euclidean metric of $R$. Let $S$ consist of all rational numbers in the open interval ($a, b$), where $a$ and $b$ are irrational. Then $S$ is ...
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1answer
44 views

Compact, extremely disconnected spaces of weight $\omega_1$

Weight of a topological space is the minimal cardinality of a basis of the topology. A space $X$ is extremely disconnected if open sets in $X$ have open closures. Is there an example in ZFC of a ...
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121 views

Is a compact simplicial complex necessarily finite?

I'm aware that a finite simplicial complex is compact, and I am wondering whether the converse is true. If we have the topological realisation of a simplicial complex (not necessarily finite), $|K|$, ...
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0answers
29 views

Compactification via embeddings and extending continuous functions

My question comes from reading Munkres' Topology, the section on Stone-Čech compactification. To find the compactification $\mathrm{Y}$ of $\mathrm{X}$, we find an embedding h, $\mathrm{h}: X ...
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1answer
25 views

Proof regarding compact $T_2$-spaces and closed continuity.

Question: Prove that any continuous function from a compact $T_2$-space onto a $T_2$-space is closed, that is, $f(F)$ is closed if $F$ is closed. Is my general reasoning correct? Any compact subset ...
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1answer
63 views

Transverse intersection in a compact manifold

Is it true that if $M$ is a compact manifold and $X,Y$ are submanifolds of $M$ which intersect transversely that the intersection $X\cap Y$ consists of finitely many points? I'm trying to understand ...
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1answer
58 views

How do i show that if every continuous function on $X$ is bounded, then $X$ is compact? [duplicate]

Let $(X,d)$ be a metric space. Assume every continuous function on $X$ is bounded. Prove that $X$ is compact. Well, i don't know which continuous function should i fix to start an ...
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2answers
143 views

Homeomorphism Compact Subsets

Are there compact subsets $A,B \subset \mathbb{R^2}$ with $A$ not homeomorphic to $B$ but $A \times [0,1]$ homeomorphic to $B \times [0,1]$?
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1answer
29 views

How can I show that this set is sequentially compact?

Let $(x_k)$ be a convergent sequence in $\mathbb{R}$ such that $(x_k) \to x$. Let $A = \{x_1, x_2, \ldots\} \cup \{x\}$. It's quite easy to show that $A$ is compact by showing that every open cover ...
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1answer
51 views

Proof of compactness of Lipschitz functions

Consider the set $\mathcal{F}$ of continuous functions on $[0;1]$ with boundary values $$ f(0)=f(1)=0 \qquad \forall f \in \mathcal{F}. $$ Define the metric $d(f,g) = \lVert f-g \rVert_\infty = ...
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1answer
53 views

Bases and open covers of a topological space

A basis of a topological space $X$ is a family of open sets ${B_i : i \in I}$ for some indexing set $I$, where any open set in $X$ can be written as the union of two or more members of ${B_i : i \in ...
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2answers
212 views

Riesz's Theorem of compactness

$\left(X,\|\cdot\|\right)$ is a normed vector space. $\textbf{Riesz's Theorem of compactness}$ says that $$ \{x \in X \colon \|x\| \leq 1 \} \ \text{compact} \ \Longleftrightarrow \ \text{Each bounded ...
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2answers
46 views

More about compact.

Any advice to solve the following problem? Let $K$ be a compact set of the real numbers. Prove that the set $|K|=\{|x|: x\in K\}$ is a compact set. Thanks a lot!
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1answer
52 views

SVD: proof of existence

I'm reading "Numerical Linear Algebra" by Lloyd Thefethen. For Singular Value Decomposition proof of existence it starts like this: "Set $\sigma_1=||A||_2$. By a compactness argument, there must be ...
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2answers
75 views

Compact sets of real number

I don't know how to prove the following statement: If $K_1, K_2$ are non empty disjoint compact subset of the real numbers, prove that there exist $k_1\in K_1$, $k_2\in K_2$ such that $|k_1-k_2|= ...
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0answers
95 views

$\ f \colon X \to X $ ,continuous function where X is compact,Hausdorff space.Show $\exists A$ st $f(A) =A$.

Suppose $\ f \colon X \to X $ is a continuous function from a compact,Hausdorff space to itself. Prove that there exists a subspace $A$ such that $f(A) =A$. I came up with an answer based on nets ...
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1answer
156 views

Properties of compact set: non-empty intersection of any system of closed subsets with finite intersection property

Let $X$ be a Hausdorff topological vector space. Let $C$ be a nonempty compact subset of $X$ and $\{C_\alpha\}_{\alpha \in I}$ be a collection of closed subsets such that $C_\alpha \subset C$ for each ...
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2answers
41 views

Is sequence convergent in subspace of compact metric space?

Problem is as follow. Let X be a compact metric space and A be a closed subset of X. Prove that every sequence in A has a convergent (note: convergent in A) subsequence. It is from my note. My ...
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1answer
35 views

Compact and convex discrete set

I am working with discrete sets but I have a doubt: is the set $\{ 0,1\}$ compact and convex? And the set $\{ 0,1\}^2=\{(0,0), (1,0), (0,1), (1,1) \}$?
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150 views

Let $p: E\to B$ be a covering map. If $B$ is compact and $p^{-1}(b)$ is finite, then $E$ is compact. [duplicate]

So I start off and assume that some $\{U_\alpha\}$ is a cover of $E$. I want to reduce this cover to a finite subcover of $E$. Since $p$ is a covering map it is also an open map, therefore ...
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1answer
22 views

Redefine a discrete compact set

I need to find twice continuously differentiable functions $g_i: \mathbb{R}^2 \rightarrow \mathbb{R}$ $i=1,\ldots,I$ such that the set $\{ 0,1\}^2=\{(0,0), (1,0), (0,1), (1,1) \}$ can be written as ...
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1answer
25 views

Define a compact and convex set through inequality constraints

I need to find twice continuously differentiable functions $g_i: \mathbb{R}^2 \rightarrow \mathbb{R}$ $i=1,...,I$ such that the set $\{ 0,1\}^2=\{(0,0), (1,0), (0,1), (1,1) \}$ can be written as $\{x ...
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1answer
50 views

$\beta \omega$ is zero dimensional and extremally disconnected

I have proved that $\beta \omega$ is zero dimensional by constructing it using ultrafilters, but I know that $\beta \omega$ is characterized up to topological equivalence by the extension property, so ...
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1answer
54 views

Is the category of these particularly nice spaces cartesian closed?

Is the category of Hausdorff, compactly generated, locally path-connected, semi-locally 1-connected spaces (and continuous maps between them) cartesian closed? If not, in what ways does it fail to be? ...
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74 views

Finite-case symmetry leads to infinite-case asymmetry

Formulas for sines or cosines of sums superficially appear to have a certain symmetry, specifically it looks as if sine and cosine play something like symmetrical roles: $$ \begin{align} & ...
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2answers
89 views

Stone-Čech compactification using ultrafilters

Let $X = \omega \cup \{ x \}$ ne the Stone-Čech compactification of $\omega$. (I am viewing $X$ as a subspace of the set of ultrafilters over $\omega$). Let, $\mathcal A$, $\mathcal B$ be two disjoint ...
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2answers
778 views

Proof that a product of two quasi-compact spaces is quasi-compact without Axiom of Choice

A topological space is called quasi-compact if every open cover of it has a finite subcover. Let $X, Y$ be quasi-compact spaces, $Z = X\times Y$. The usual proof that $Z$ is quasi-compact uses a ...
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3answers
66 views

Developing an intuition for compact and open sets

I'm having trouble picturing what compact sets and open sets actually are. Open and closed intervals make enough sense to me, but for whatever reason, moving to the next level of abstraction is ...
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56 views

Weak sequential compactness in a reflexive space

Let $\{X, \| \cdot \|\}$ be a normed space, $B$ is the unit ball of $X$. If $\{X, \| \cdot \|\}$ is reflexive, then is $B$ weakly sequentially compact? If it's not true, are there any counterexamples ...
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222 views

Can we prove that a bounded closed subset of $\mathbb R^n$ is compact without Axiom of Choice?

Can we prove that a bounded closed subset of $\mathbb R^n(n \ge 1)$ is compact without using Axiom of Choice? This is a related question which was closed.
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Onto continuous function on a compact metric space is isometry. [duplicate]

Let $K$ be a compact metric space with metric $d$ and suppose $f:K\rightarrow K$ is continuous and surjective (onto), and satisfies $d(f(x),f(y))\leq d(x,y),\,\forall x,y\in K$. How can we prove that ...
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24 views

Surjective function on a compact metric space [duplicate]

Assume $f:K\rightarrow K$, is surjective and $K$ is a compact metric space and we have $d(f(x),f(y))\leq d(x,y)\, \forall x,y\in K$. How can I prove that $d(f(x),f(y))= d(x,y)\, \forall x,y\in K$? ...
4
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1answer
81 views

Lindelöf if and only if every collection with the countable intersection property has non-empty intersection of closures

I am trying to study for my topology exam, and my professor recommended this question from the text (Munkres's Topology (2nd edition), Section 37 question 2): A collection $\mathcal{A}$ of subsets of ...
5
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1answer
101 views

Compact topological space not having Countable Basis?

Does there exist a compact topological space not having countable basis? I have constructed a product space from uncountably many unit intervals $[0,1]$, endowed with the product topology. ...
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55 views

Why is proof of the [topological] closed graph theorem incorrect?

Specifically, the closed graph theorem I am referring to is: Let $f : X \rightarrow Y$ exist and $Y$ be compact and Hausdorff. Then $f$ is continuous if and only if the graph of $f$ denoted by $G_f = ...
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1answer
1k views

A proof about $F_\sigma$, $\sigma$-compact sets, and subsets of the irrationals

I've been looking at a proof that shows the following result. $\mathbb{P}$ is the set of irrational numbers, $\mathbb{Q}$ the rationals, and $\mathbb{R}$ the reals. The following conditions are ...
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1answer
50 views

Compactness in Complete Lattices

Let $(L,\leq)$ be a complete lattice. We say an $a\in L$ is compact in $L$ iff for $\forall A\subset L$ such that $a\leq \bigvee A$, there exists a finite subset $A'\subset A$, sucth that ...
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1answer
79 views

A closed bounded set having exactly one accumulation point has the covering property

I need to show that a closed, bounded set having exactly one accumulation point has the covering property. A set has the covering property if any open cover of it has a finite subcover. Since the ...
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2answers
65 views

Limit point compactness

Let $(X, d)$ be a metric space and $A ⊆ X$. If $A$ is limit point compact, show that $A$ is closed. My thoughts: The definition of limit point compactness is that for each infinite subset of $A$, it ...
3
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1answer
54 views

About the Stone-Čech universal property

There's something I am missing here and I don't know what it is. I understand that the Stone-Čech compactification of $X$ satisfies the property that for every continuous map $f: X \rightarrow K$ ...
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99 views

understand proof of compactness in product topology

I am trying to understand the following reasoning. Call $\mathcal{F_\lambda}$ the set of functions $a:\mathbb{N} \to \mathbb{R}$ for which $Na(i) := \sum_{j \in \mathbb{N}} n_{ij} a(j)\leq \lambda ...
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1answer
52 views

Constructing a Set with Connected Interior

Suppose that $K\subset\mathbb C$ is a compact set with non-empty interior and suppose that $a\in\operatorname{int} K$. I want to construct a set $M$ with the following properties: $M\subseteq K$; ...
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15 views

How to use a base to prove something is sequentially compact.

I know this is not very specific but I'm studying for a topology exam and this is one of the things I need to know how to do. I know that part of the process is showing it converges. I was hoping ...