Question

Prove that closed bounded set contains supremum

score 492 · Answer 1 · 2026-03-29 14:28:14

492

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Prove that closed bounded set contains supremum

Published on 29 Mar 2026 - 14:28

score 165 · Answer 2 · 2026-03-25 11:15:04

165

Views

Can someone find the error in my proof that if X is compact then it's a manifold?

Published on 25 Mar 2026 - 11:15

score 390 · Answer 3 · 2019-11-18 05:54:46

390

Views

Compactness of the subspace (Relative topology)

Published on 18 Nov 2019 - 5:54

score 126 · Answer 4 · 2026-03-31 16:21:40

126

Views

Only base in definition of compactness - do we need AC?

Published on 31 Mar 2026 - 16:21

score 217 · Answer 5 · 2026-03-31 17:58:49

217

Views

"Indexed" version of compactness and Axiom of Choice

Published on 31 Mar 2026 - 17:58

score 271 · Answer 6 · 2019-11-18 20:43:21

271

Views

A sequentially compact space is countably compact.

Published on 18 Nov 2019 - 20:43

score 30 · Answer 7 · 2019-11-19 01:21:15

30

Views

$\overline{F^c \cap K^c}$

Published on 19 Nov 2019 - 1:21

score 1.2k · Answer 8 · 2019-11-19 23:28:20

1.2k

Views

Show that every locally compact Hausdorff space is regular.

Published on 19 Nov 2019 - 23:28

score 92 · Answer 9 · 2026-03-26 19:02:48

92

Views

$T_{a}(\text{ball}(\ell^p))$ is compact, where $T_a\colon\ell^p\to\ell^p$, $(T_a x)_{n}:=a_{n}x_{n}$ and $\mathbb{C}\ni a_n\to0$ as $n\to\infty$

Published on 26 Mar 2026 - 19:02

score 57 · Answer 10 · 2026-02-23 02:57:47

57

Views

Weak compactness in measures implies compactness in the underlying metric space via the Dirac's delta

Published on 23 Feb 2026 - 2:57

score 120 · Answer 11 · 2019-11-21 14:41:32

120

Views

Application of Arzela-Ascoli

Published on 21 Nov 2019 - 14:41

score 132 · Answer 12 · 2019-11-21 18:52:09

Let $U\subseteq X$ be an open set such that $\bigcap\limits_{n=1}^\infty A_n\subseteq U$. Show that $A_m\subseteq U$ for some $m\geq 1$

score 176 · Answer 13 · 2019-11-21 23:26:33

176

Views

Must every infinite subshift contain a non-periodic point?

Published on 21 Nov 2019 - 23:26

score 42 · Answer 14 · 2026-03-28 05:22:11

42

Views

Relaxing homeomorphism to compact domain to preserve fixed point property.

Published on 28 Mar 2026 - 5:22

score 191 · Answer 15 · 2026-03-31 17:54:44

191

Views

Is the axiom of choice needed in proving that metric spaces in which every infinite subset has a limit point are compact?

Published on 31 Mar 2026 - 17:54

List Question

Prove that closed bounded set contains supremum

Can someone find the error in my proof that if X is compact then it's a manifold?

Compactness of the subspace (Relative topology)

Only base in definition of compactness - do we need AC?

"Indexed" version of compactness and Axiom of Choice

A sequentially compact space is countably compact.

$\overline{F^c \cap K^c}$

Show that every locally compact Hausdorff space is regular.

$T_{a}(\text{ball}(\ell^p))$ is compact, where $T_a\colon\ell^p\to\ell^p$, $(T_a x)_{n}:=a_{n}x_{n}$ and $\mathbb{C}\ni a_n\to0$ as $n\to\infty$

Weak compactness in measures implies compactness in the underlying metric space via the Dirac's delta

Application of Arzela-Ascoli

Let $U\subseteq X$ be an open set such that $\bigcap\limits_{n=1}^\infty A_n\subseteq U$. Show that $A_m\subseteq U$ for some $m\geq 1$

Must every infinite subshift contain a non-periodic point?

Relaxing homeomorphism to compact domain to preserve fixed point property.

Is the axiom of choice needed in proving that metric spaces in which every infinite subset has a limit point are compact?

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