Question

In a compact metric space, every sequence has a convergent subsequence

score 1.7k · Answer 1 · 2026-04-15 11:24:37

1.7k

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In a compact metric space, every sequence has a convergent subsequence

Published on 15 Apr 2026 - 11:24

score 594 · Answer 2 · 2026-04-14 08:16:11

594

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Reference for convergence properties of the Hausdorff distance

Published on 14 Apr 2026 - 8:16

score 99 · Answer 3 · 2026-04-14 03:26:02

99

Views

No direct proofs of "if $ f: (X, d_X) \to (Y, d_Y)$ is continuous and $X$ is compact then $f$ is uniformly continuous."

Published on 14 Apr 2026 - 3:26

score 888 · Answer 4 · 2026-04-10 16:50:56

888

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Continuous function vs Uniformly continuous function

Published on 10 Apr 2026 - 16:50

score 17.6k · Answer 5 · 2026-04-13 12:17:55

17.6k

Views

For continuous functions, preimage of open set is open.

Published on 13 Apr 2026 - 12:17

score 475 · Answer 6 · 2026-04-14 12:51:17

475

Views

Does every non-compact bounded metric space support an equivalent metric in which it is unbounded?

Published on 14 Apr 2026 - 12:51

score 129 · Answer 7 · 2026-04-15 10:57:57

129

Views

Why is the triangle inequality property of a metric space important?

Published on 15 Apr 2026 - 10:57

score 1k · Answer 8 · 2026-04-16 03:46:12

1k

Views

Prove that the space of sequences under this metric is complete and compact.

Published on 16 Apr 2026 - 3:46

score 263 · Answer 9 · 2026-04-10 23:16:36

263

Views

Assumptions of Kolmogorov extension theorem

Published on 10 Apr 2026 - 23:16

score 822 · Answer 10 · 2026-04-13 21:12:41

822

Views

Does uniform convergence depends on the metric?

Published on 13 Apr 2026 - 21:12

score 187 · Answer 11 · 2026-04-15 05:04:08

187

Views

Metric on natural numbers united with infinity

Published on 15 Apr 2026 - 5:04

score 1.4k · Answer 12 · 2026-04-10 22:04:46

1.4k

Views

triangle inequality for a metric space

Published on 10 Apr 2026 - 22:04

score 1k · Answer 13 · 2026-04-18 11:55:53

1k

Views

$d$ is a metric space on $X\not=\{0\}$, obtained from a norm. $d'(x,y)=d(x,y)+1$. Show $d'$ cannot be obtained from a norm.

Published on 18 Apr 2026 - 11:55

score 60 · Answer 14 · 2026-04-15 22:22:17

60

Views

A map $f:([a,b], |\cdot|) \to ([c,d], |\cdot|)$ is an isometry if and only if $d-c = b-a$.

Published on 15 Apr 2026 - 22:22

score 506 · Answer 15 · 2026-04-15 15:30:14

506

Views

$\epsilon$-isometry of a compact metric space is $\epsilon$-surjective

Published on 15 Apr 2026 - 15:30

List Question

In a compact metric space, every sequence has a convergent subsequence

Reference for convergence properties of the Hausdorff distance

No direct proofs of "if $ f: (X, d_X) \to (Y, d_Y)$ is continuous and $X$ is compact then $f$ is uniformly continuous."

Continuous function vs Uniformly continuous function

For continuous functions, preimage of open set is open.

Does every non-compact bounded metric space support an equivalent metric in which it is unbounded?

Why is the triangle inequality property of a metric space important?

Prove that the space of sequences under this metric is complete and compact.

Assumptions of Kolmogorov extension theorem

Does uniform convergence depends on the metric?

Metric on natural numbers united with infinity

triangle inequality for a metric space

$d$ is a metric space on $X\not=\{0\}$, obtained from a norm. $d'(x,y)=d(x,y)+1$. Show $d'$ cannot be obtained from a norm.

A map $f:([a,b], |\cdot|) \to ([c,d], |\cdot|)$ is an isometry if and only if $d-c = b-a$.

$\epsilon$-isometry of a compact metric space is $\epsilon$-surjective

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