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 · 2026-04-15 04:07:35

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 15 Apr 2026 - 4:07

score 112 · Answer 2 · 2026-05-10 12:38:27

112

Views

On a comparison of connected components of Noetherian topological spaces

Published on 10 May 2026 - 12:38

score 160 · Answer 3 · 2026-05-11 16:39:30

160

Views

Countable union covers a second countable, compact Hausdorff space

Published on 11 May 2026 - 16:39

score 4.3k · Answer 4 · 2026-04-16 18:41:38

4.3k

Views

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

Published on 16 Apr 2026 - 18:41

score 49 · Answer 5 · 2026-04-15 01:37:59

49

Views

$T_2\implies T_4$ without compactness hypothesis?

Published on 15 Apr 2026 - 1:37

score 53 · Answer 6 · 2026-05-11 08:30:47

53

Views

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

Published on 11 May 2026 - 8:30

score 77 · Answer 7 · 2026-05-07 15:40:55

77

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 07 May 2026 - 15:40

score 2.2k · Answer 8 · 2026-05-10 14:42:55

2.2k

Views

Compactness gives Second countable space??

Published on 10 May 2026 - 14:42

score 265 · Answer 9 · 2026-05-10 14:32:35

265

Views

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

Published on 10 May 2026 - 14:32

score 50 · Answer 10 · 2026-05-10 15:00:41

50

Views

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

Published on 10 May 2026 - 15:00

score 326 · Answer 11 · 2026-05-10 12:37:45

326

Views

Is there a closed totally disconnected space meeting every component?

Published on 10 May 2026 - 12:37

score 234 · Answer 12 · 2026-04-14 00:01:46

234

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 14 Apr 2026 - 0:01

score 205 · Answer 13 · 2026-04-12 17:05:30

205

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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 12 Apr 2026 - 17:05

score 1.6k · Answer 14 · 2026-04-16 06:13:52

1.6k

Views

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

Published on 16 Apr 2026 - 6:13

score 335 · Answer 15 · 2026-05-10 12:55:11

335

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 10 May 2026 - 12:55

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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