Question

Distance between subsets of metric spaces

score 43 · Answer 1 · 2019-10-12 12:55:36

43

Views

Distance between subsets of metric spaces

Published on 12 Oct 2019 - 12:55

score 53 · Answer 2 · 2026-04-01 01:15:47

53

Views

Why $\{(x,y)\in\Bbb{R}^2 :\, y=x\sin \frac1x\}\cup\{(0,0)\}$ is connected but not compact?

Published on 01 Apr 2026 - 1:15

score 491 · Answer 3 · 2026-03-28 23:18:50

491

Views

Proof of the continuous function on a compact set is uniform continuous

Published on 28 Mar 2026 - 23:18

score 22 · Answer 4 · 2019-10-15 12:01:34

22

Views

Are these sets bounded and compact or not?

Published on 15 Oct 2019 - 12:01

score 149 · Answer 5 · 2026-03-25 17:52:57

149

Views

Compact-Hausdorff space in general topology

Published on 25 Mar 2026 - 17:52

score 68 · Answer 6 · 2019-10-16 23:35:15

68

Views

I am trying to reach a contradiction here.

Published on 16 Oct 2019 - 23:35

score 192 · Answer 7 · 2019-10-17 14:15:20

192

Views

Proper map from continuous if it maps compact sets to compact sets

Published on 17 Oct 2019 - 14:15

score 673 · Answer 8 · 2019-10-17 21:03:08

673

Views

Can I apply Urysohn's lemma on a locally compact Hausdorff space?

Published on 17 Oct 2019 - 21:03

score 74 · Answer 9 · 2026-03-25 22:17:44

74

Views

Proving $X$ is compact using a function $f\colon X \to Y$?

Published on 25 Mar 2026 - 22:17

score 71 · Answer 10 · 2019-10-22 04:32:24

71

Views

$\{x_n\}_{n=1}^\infty$, $\{y_n\}_{n=1}^\infty$ $\subset M$ (compact) then $\exists$ $\{x_{n_k}\}, \{y_{n_k}\}$ s.t. they both converge to same point.

Published on 22 Oct 2019 - 4:32

score 870 · Answer 11 · 2019-10-22 12:09:19

870

Views

Why define a compact set through subsequences?

Published on 22 Oct 2019 - 12:09

score 500 · Answer 12 · 2019-10-23 08:52:37

500

Views

Every closed and bounded subset of $(R^n,||\cdot||_p)$ is compact $1\leq p \leq \infty$

Published on 23 Oct 2019 - 8:52

score 45 · Answer 13 · 2019-10-23 09:13:27

45

Views

Show that $Y=\{e_n | n\in N\}\subset (\ell_p,d_p),\; 1\leq p< \infty$ is neither totally bounded nor compact

Published on 23 Oct 2019 - 9:13

score 669 · Answer 14 · 2019-10-23 09:46:17

669

Views

Every subset $Y$ of a totally bounded metric space $(X,d)$ is also totally bounded

Published on 23 Oct 2019 - 9:46

score 729 · Answer 15 · 2019-10-23 10:35:15

729

Views

$(X,d)$ Complete and totally bounded $\iff $ $(X,d)$ is compact

Published on 23 Oct 2019 - 10:35

List Question

Distance between subsets of metric spaces

Why $\{(x,y)\in\Bbb{R}^2 :\, y=x\sin \frac1x\}\cup\{(0,0)\}$ is connected but not compact?

Proof of the continuous function on a compact set is uniform continuous

Are these sets bounded and compact or not?

Compact-Hausdorff space in general topology

I am trying to reach a contradiction here.

Proper map from continuous if it maps compact sets to compact sets

Can I apply Urysohn's lemma on a locally compact Hausdorff space?

Proving $X$ is compact using a function $f\colon X \to Y$?

$\{x_n\}_{n=1}^\infty$, $\{y_n\}_{n=1}^\infty$ $\subset M$ (compact) then $\exists$ $\{x_{n_k}\}, \{y_{n_k}\}$ s.t. they both converge to same point.

Why define a compact set through subsequences?

Every closed and bounded subset of $(R^n,||\cdot||_p)$ is compact $1\leq p \leq \infty$

Show that $Y=\{e_n | n\in N\}\subset (\ell_p,d_p),\; 1\leq p< \infty$ is neither totally bounded nor compact

Every subset $Y$ of a totally bounded metric space $(X,d)$ is also totally bounded

$(X,d)$ Complete and totally bounded $\iff $ $(X,d)$ is compact

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