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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Every minimal KC space is compact

A space $(X,\tau )$ is said to be minimal $KC$ if $(X,\tau)$ is $KC$ but no topology on $X$ which is strictly smaller than $\tau$ is $KC$. Theorem : Every minimal KC space is compact. Proof. ...
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Normalizer of an open subgroup of a compact group

For some reason I cannot figure out the following. Suppose $G$ is a compact, Hausdorff and totally disconnected (but I think compact should suffice) group and $U$ an open subgroup. Why is the ...
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64 views

locally compact and compatification

As we know a topological space $X$ is said to be locally compact at a point $ x \in X $ if $x$ has a compact neighbourhood in $X$. $X$ is called locally compact if it is locally compact at every ...
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103 views

About $ \{ x \in[0,1]^{\omega_1}:|\{\alpha<\omega_{1} :x(\alpha)\ne 0 \}|\le\omega \}$

Take $X$ a Tychonoff product $[0,1]^{\omega_1}$ and as $Y$ the $\Sigma$-product $$ \{ x ∈[0,1]^{\omega_1}:|\{\alpha<\omega_{1} :x(\alpha)\ne 0 \}|\le\omega \}\;.$$ The space $X$ is compact by ...
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1answer
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A sequential minimal KC-space is compact

A space $(X,\tau )$ is said to be minimal KC , if $(X,\tau )$ is KC but no topology on X which is strictly smaller than $ \tau$ will be KC Theorem : A sequential minimal KC-space is compact. ...
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2answers
165 views

If $A$ is compact and $B$ is Lindelöf space , will be $A \cup B$ Lindelöf

I have 2 different questions: As we know a space Y is Lindelöf if each open covering contains a countable subcovering. (1) :If $A$ is compact and $B$ is Lindelöf space , will be $A \cup B$ ...
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1answer
52 views

Accumulation point

Accumulation point: Let $ \tau$ be a topological on $ ‎\mathbb{N}‎ $‎‎ such that is generated by $\{1,2\}, \{3,4\},\{5,6\}.... $. Let $A$ be non-empty of $ ‎\mathbb{N}‎ $‎‎ and $ n_{0} \in A$. If $ ...
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389 views

The proof of Cantor's Intersection Theorem on nested compact sets

The book "Metric Spaces" by Babu Ram says this about the proof of Cantor's Intersection Theorem: Create nested intervals $F_{n+1}\subset F_n$ such that $\lim_{n\to\infty}\text{diameter}(F_n)=0$. ...
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1answer
59 views

Prove that if $\mathfrak{J}\subset\mathfrak{J'}$ and $(X,\mathfrak{J'})$ is compact, then $(X,\mathfrak{J})$ is compact.

I'm reading Intro to Topology by Mendelson. The problem statement is in the title. My attempt at the proof is as follows: Let $\lbrace U_\alpha\rbrace_{\alpha\in I}$ be an open covering of $X$ such ...
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2answers
75 views

infinite regular cardinal

Let $(X,\tau)$ be a KC non-compact space. Then there is a discrete subset $D \subseteq X$ such that $\overline D$ is not compact. Furthermore there is an ultrafilter $F$ in $X$ such that $ D \in F $ ...
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31 views

ultrafilter and $KC$-minimal space

By definition in a $KC$ space every compact set is closed. A space $(X,‎\tau)‎‎‎$ is $KC$-minimal if $(X,‎\tau )‎‎‎$ is a $KC$ space and there is no $KC$ space $(X,\sigma)$ such that ...
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123 views

Prove $X$ is compact iff for each family $\{ F_\alpha\}_{\alpha\in I}$ of closed subsets of $X$ with the FIP, we have $\cap F_\alpha\neq\varnothing$.

I'm reading Intro to Topology from Mendelson. The entire problem statement is, Prove that $X$ is compact if and only if for each family $\lbrace F_\alpha\rbrace_{\alpha\in I}$ of closed subsets of ...
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1answer
29 views

A hereditarily Lindelöf $KC$-space $( X,τ )$ is Katětov-$KC$ if and only if there is a weaker sequential $US$ topology $σ⊂τ

A space $( X,τ )$ is said to be Katětov $ KC $ if there is a topology $ σ⊂τ$ such that $( X,σ )$ is minimal $ KC $. The notion of strongly KC-spaces, that is, those spaces in which every ...
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36 views

Let $X$ be a$KC$ space. Then $X^{*}$ is $KC$ iff $X$ is a $K$ -space

The bellow theorem exist in " Between $T_{1} $ and $T_{2}$ " by " Albert Wilansky ." Let $ (‎ ‎X‎^{*},‎\tau‎^{*} ‎)‎ $ be topological space one- point compatification of $ ( X. \tau)$. A topological ...
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1answer
92 views

Prove that $x\not\in P$ then there exists a neighborhood $O$ of $x$ such that $O\cap P=\varnothing$. Certain definition on $P$.

I'm reading Intro to Topology by Mendelson. The problem statement is, Let $\{U_\alpha\}_{\alpha\in I}$ be an open covering of $[0,1]$. Define a subset $P$ of $[0,1]$ as follows: $x\in P$ if ...
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1answer
111 views

complete accumulation point

Let $(X, ‎\tau‎ )$ be a topological space, which is not compact. Then there is$ C ⊆ X$, such that $ C$ has no complete accumulation point. Proof. A space $X$ is not compact, hence there is a ...
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1answer
34 views

An infinite minimal strongly KC-space possesses a non-trivial

The notion of strongly KC-spaces mean spaces in which every countably compact subset is closed. a space $(X,‎\tau‎ )$ is said to be minimal strongly KC if $(X,‎\tau‎ )$ is strongly KC but no ...
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1answer
153 views

Prove that $x\in P$, then there exists a neighborhood $O$ of $x$ such that $O\subset X$, $P$ is defined in a particular way.

I'm reading Intro to Topology by Mendelson. This is the first section on compact topological spaces. The complete question is, Let $\{U_\alpha\}_{\alpha\in I}$ be an open covering of $[0,1]$. ...
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1answer
70 views

For a holomorphic function, does Newton's difference quotient converge uniformly on compact sets?

Let $K\subset A\subset C$, where $K$ is compact, $A$ is open, and $C$ is the complex numbers. Let $f:C\rightarrow C$ be holomorphic on $A$ (holomorphic = complex derivative exists for all points in ...
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1answer
337 views

Is this set compact? Connected?

Is this set compact? Connected? $S=\{(x,y,z)\in\mathbb{R}^3:z=x^2+y^2+1\}$ for $z\le 1$ this set is not defined, but for $z>1$ we are getting circles! I imagined this as a bunch of ...
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Compactness used to get a covering by special smaller balls

Suppose $(X,d)$ is a compact metric space. Suppose we have a set $A \subseteq X$ such that the set of open $\epsilon$-balls around the points of $A$ cover $X$. I've read that "By compactness, there ...
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1answer
198 views

Bounded sets of isolated points in compact metric spaces

Context and definitions: Say $(X,d)$ is a compact metric space, with $f: X \rightarrow X$ continuous. For each $n \in \mathbb{N}$, the metric $d_{n}(x,y) = \max_{0 \leq k \leq ...
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$ KC $ spaces imply $ US $ spaces , but vise versa is false.

In the $ US $ space , each convergent sequence has unique limit. In the $ KC $ space , every compact subset is closed. It easy to show that $ KC $ spaces imply $ US $ spaces. The ...
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90 views

if $ X$ is a countable, compact $ T_{1} $ space and $ A ‎\subseteq‎‎‎‎ X $ then either $A$ is compact or…

Theorem: if $ X$ is a countable, compact $ T_{1} $ space and $ A ‎\subseteq‎‎‎‎ X $ then either $A$ is compact or there is a sequence in $A$ converging to point of $ X- A $. proof: Suppose ...
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1answer
59 views

Question on compactness of $\Delta(X)$ with the weak$^{\ast}$ topology for compact $X$

Let $X$ be a metric space. $\Delta(X)$ is the set of all probability Borel measures on $X$(which is regular, of course).$\Delta(X)$ is endowed with weak$^{\ast}$ topology for which the mapping $\mu ...
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1answer
993 views

If every convergent subsequece converges to the same limit then the sequence converges

I came across this question: In a compact metric space $(X,d)$ if every convergent subsequence of a sequence converges to the same limit, say $l$, then the original sequence also converges to $l$.
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Let $ X $ be compact and let $f: X \rightarrow Y$ be a local homeomorphism. Show that for any point $y \in Y$, $f^{-1}(y)$ is a finite set.

PROBLEM: Let $ X $ be compact and let $f: X \rightarrow Y$ be a local homeomorphism. Show that for any point $y \in Y$, $f^{-1}(y)$ is a finite set. I was trying out this problem from Massey, but got ...
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1answer
196 views

Show that a subset of $\mathbb{C}$ is compact.

Let $\Omega$ be a bounded set in $\mathbb{R}^n$ and $s>0$. Set $$\Omega_s := \bigcup_{a\in\Omega} {B(a,s)}$$ where $$B(a,s) = \left\{ z \in \mathbb{C}^n : \sup_{1 \leq i \leq n} {|z_i - a_i|} ...
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1answer
228 views

Every bounded subset of $R^k$ lies in a compact subset of $R^k$

P: Every bounded subset of R$^k$ lies in a compact subset of R$^k$ (p52, Rudin) How is the above true? I know that, by theorem, a set $E$ in $R^k$ is closed and bounded if and only if $E$ is ...
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0answers
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Every Borel set is the union of an increasing sequence of Bounded Borel sets?

I am currently working with the book by Halmos, and i can't quite get past this one. It states that: "Every Borel set can be written as an increasing sequence of Bounded Borel sets" In this case $X$ ...
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1answer
64 views

minimal KC and (strongly) KC

If P is a topological property, then a space (X, τ) is said to be minimal P (respectively, maximal) if (X,τ) has property P but no topology on X which is strictly smaller (respectively, strictly ...
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87 views

topological space

Let $( X,\tau )$ be a $T_1$ topological space. Let $D = \{ d_n : n \in \omega \}$ be a countably infinite closed discrete subspace of $X$. Fix $P \in X$ and let $F \in \beta\omega- \omega$ be an ...
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126 views

Orthogonal chords of compact sets

For any compact set on a plane say C does there always exist a chord in C such that its end points are orthogonal to the boundary of C (assumed smooth)
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76 views

Prove that $H$ is compact $\iff$ every cover $\{E_{\alpha}\}_{\alpha \in A}$ has a finite subcovering.

Let $H \subseteq \Bbb R^n$. Prove that $H$ is compact $\iff$ every cover $\{E_{\alpha}\}_{\alpha \in A}$ where $E_{\alpha}$'s are relatively open in $H$ has a finite subcovering. $\bf{Solution \ ...
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70 views

Prove that $H$ is a finite set.

Let $H$ be compact in $\Bbb R^n$ Also assume that for every $x\in H$ there is an $r=r(x)$ such that $B_r(x)\cap H=\{x\} $ Prove that $H$ is a finite set. Solution: Since $H$ is compact, ...
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2answers
79 views

Compact and countability axioms!

I wondering which countability axioms compact imply in arbitrary topological spaces. I'm using Greene/Gamelin 2nd ed. And they list separable, 2nd-coutable, first-countable and Lindelöf. Clearly ...
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1answer
154 views

I don't understand this remark regarding Weierstrass' Theorem (Ahlfors' Complex Analysis)

In Ahlfors' text of Complex Analysis, chapter 5 theorem 1, he proves the following: Theorem I. Suppose that $f_n(z)$ is analytic in the region $\Omega_n$, and that the sequence $\{f_n(z)\}$ ...
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116 views

The Stone-Čech compactification of a space by the maximal ideals of the ring of bounded continuous functions from the space to $\mathbb{R}$

There is a claim that for any completely regular space, the maximal ideals of the ring of bounded continuous functions from $X$ to $\mathbb{R}$ forms the Stone-Čech compactification of $X$. How is the ...
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1answer
44 views

$X$ is a complete metric space, $Y$ is compact. $X \times Y$ is Baire?

Requesting a hint or solution. X is a complete metric space and Y is a compact hausdorff space. Trying to show that $X \times Y$ is a Baire space.
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2answers
513 views

If $E$ is an infinite subset of a compact set $K$, then $E$ has a limit point in $K$

The below is the proof for the theorem that if $E$ is an infinite subset of a compact set $K$, then $E$ has a limit point in $K$. The only part I can't accept in the proof is the last sentence. So, ...
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2answers
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True Or not: Compact iff every continuous function is bounded [duplicate]

Let $X$ be a topological space. My question is: If $f:X\to \mathbb{R}$ is bounded for all such continuous $f$, then is $X$ compact. Is is really? If $X$ is the subset of $\mathbb{R}^d$, then it is ...
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1answer
338 views

Can compacts on a real line behave “paradoxically” with respect to unions, intersections, and translation? What about other topological groups?

I have the following question i cannot answer myself. Consider two compacts $A$ and $B$ on the real line $\mathbb R$. Let $A'$ be a translation of $A$ and $B'$ a translation of $B$: $A' = A + a$, ...
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1answer
568 views

Compactly supported continuous function is uniformly continuous

Let $f:\mathbb R \rightarrow \mathbb R$ be continuous and compactly supported. How can I prove that $f$ is uniformly continuous ? I was trying to prove it by contradiction but get stuck. My attempt ...
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3answers
351 views

Baby Rudin 2.26 Infinite subsets with limit points implies compactness

Having some trouble with this question. Let $X$ be a metric space in which every infinite subset has a limit point. Prove that $X$ is compact. Hint: By Exercises 23 and 24, $X$ has a ...
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1answer
187 views

Is there a non-locally compact Hausdorff space in which all infinite compact sets (of which there is at least one) have uncountable interiors?

Here is the background material from which I am working: The Cantor set is an uncountable compact Hausdorff subspace of $\mathbb{R}$ with empty interior. In a locally compact Hausdorff space with no ...
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537 views

Prove that K, a subset of R, is compact

I thought this question is quite easy. My proof is that In R, the subset is compact if and only if is closed and bounded. K is bounded since d(1/2, r) < 1 for any r in K, and it is closed since its ...
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1answer
111 views

Is there a non-locally compact Hausdorff space in which all infinite compact sets (of which there is at least one) have nonempty interior?

Here is the background material from which I am working: The Cantor set is an uncountable compact Hausdorff space with empty interior. In a locally compact Hausdorff space, each countable set has ...
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1answer
238 views

Show that $\mathbb R$ is countably compact

Show that $\mathbb R$ is countably compact, that is that any cover has an at most countable subcover. I tried to use the fact that every open set in R is a union of intervals with rational ...
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2answers
64 views

Are one point compactifications topologically invariant?

More precisely: are one point compactifications $X^+$ and $Y^+$ of homeomorphic locally compact Hausdorff spaces $X$ and $Y$ again homeomorphic? Kindly appreciated, Aris
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1answer
197 views

Lebesgue's criterion for Riemann integrability of bounded real valued functions defined on compact metric spaces

Let $(X, d)$ be a compact metric space, and let $S$ be the algebra of sets generated by the open and closed balls of $X$. Suppose we have a pre-measure defined on $S$ such that the measure of each ...