Question

A set $S \subset \mathbb{R}$ has no cluster points iff $S \cap [-n,n]$ is a finite set for each $n \geq 1$

score 1k · Answer 1 · 2018-02-10 17:46:44

1k

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A set $S \subset \mathbb{R}$ has no cluster points iff $S \cap [-n,n]$ is a finite set for each $n \geq 1$

Published on 10 Feb 2018 - 17:46

score 110 · Answer 2 · 2026-03-25 11:54:07

110

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On a comparison of connected components of Noetherian topological spaces

Published on 25 Mar 2026 - 11:54

score 151 · Answer 3 · 2026-03-24 06:42:59

151

Views

Countable union covers a second countable, compact Hausdorff space

Published on 24 Mar 2026 - 6:42

score 4.3k · Answer 4 · 2026-03-26 18:57:46

4.3k

Views

Nested sequence of non-empty compact subsets - intersection differs from empty set

Published on 26 Mar 2026 - 18:57

score 48 · Answer 5 · 2026-03-26 18:56:56

48

Views

$T_2\implies T_4$ without compactness hypothesis?

Published on 26 Mar 2026 - 18:56

score 50 · Answer 6 · 2026-03-25 18:31:21

50

Views

From $X=\bigcup_{j=1}^n O_{y_j}$ to $\bigcap_{j=1}^n O_{y_j} \supset X$?

Published on 25 Mar 2026 - 18:31

score 67 · Answer 7 · 2026-03-17 08:34:26

67

Views

Show that for an abelian countable group $G$ there exists a compact path connected subspace $K ⊆ \Bbb R^4$ such that $H_1(K)$ isomorphic to $G$

Published on 17 Mar 2026 - 8:34

score 2.2k · Answer 8 · 2026-03-22 16:41:43

2.2k

Views

Compactness gives Second countable space??

Published on 22 Mar 2026 - 16:41

score 262 · Answer 9 · 2026-03-25 01:20:29

262

Views

Is the space of all convergent sequences compact in the space of all bounded sequences($l^{\infty})$?

Published on 25 Mar 2026 - 1:20

score 47 · Answer 10 · 2026-03-24 23:45:51

47

Views

Compact group acting on compact set s.t. orbit of net converges

Published on 24 Mar 2026 - 23:45

score 322 · Answer 11 · 2026-03-25 11:53:13

322

Views

Is there a closed totally disconnected space meeting every component?

Published on 25 Mar 2026 - 11:53

score 231 · Answer 12 · 2018-02-16 22:26:39

231

Views

If A is a proper closed subalgebra of $C(X)$ ($X$ compact) and A $\subseteq I_{x_0}$ then $A = I_{x_0}$

Published on 16 Feb 2018 - 22:26

score 203 · Answer 13 · 2018-02-18 14:21:53

203

Views

Let $V$ be a real normed finite dimensional vector space. Is it true that $A \subseteq V$ compact iff $A$ is closed and bounded?

Published on 18 Feb 2018 - 14:21

score 1.6k · Answer 14 · 2018-02-19 15:01:53

1.6k

Views

Intersection of a family of compact sets having finite intersection property in a Hausdorff space

Published on 19 Feb 2018 - 15:01

score 331 · Answer 15 · 2026-03-25 04:43:29

331

Views

$X_\alpha$ locally compact and all but a finite number of factors are compact $ \implies\prod_\alpha X_\alpha$ is locally compact

Published on 25 Mar 2026 - 4:43

List Question

A set $S \subset \mathbb{R}$ has no cluster points iff $S \cap [-n,n]$ is a finite set for each $n \geq 1$

On a comparison of connected components of Noetherian topological spaces

Countable union covers a second countable, compact Hausdorff space

Nested sequence of non-empty compact subsets - intersection differs from empty set

$T_2\implies T_4$ without compactness hypothesis?

From $X=\bigcup_{j=1}^n O_{y_j}$ to $\bigcap_{j=1}^n O_{y_j} \supset X$?

Show that for an abelian countable group $G$ there exists a compact path connected subspace $K ⊆ \Bbb R^4$ such that $H_1(K)$ isomorphic to $G$

Compactness gives Second countable space??

Is the space of all convergent sequences compact in the space of all bounded sequences($l^{\infty})$?

Compact group acting on compact set s.t. orbit of net converges

Is there a closed totally disconnected space meeting every component?

If A is a proper closed subalgebra of $C(X)$ ($X$ compact) and A $\subseteq I_{x_0}$ then $A = I_{x_0}$

Let $V$ be a real normed finite dimensional vector space. Is it true that $A \subseteq V$ compact iff $A$ is closed and bounded?

Intersection of a family of compact sets having finite intersection property in a Hausdorff space

$X_\alpha$ locally compact and all but a finite number of factors are compact $ \implies\prod_\alpha X_\alpha$ is locally compact

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