Question

Why aren't the rationals a compact subset of $\mathbb{R}$?

score 1.5k · Answer 1 · 2026-04-11 09:15:15

1.5k

Views

Why aren't the rationals a compact subset of $\mathbb{R}$?

Published on 11 Apr 2026 - 9:15

score 32 · Answer 2 · 2026-04-14 02:52:29

32

Views

Show that a set is weakly compact nonseparable.

Published on 14 Apr 2026 - 2:52

score 242 · Answer 3 · 2026-04-13 04:41:43

242

Views

Show that $T^$ is weak$^$-to-weak$^*$ continuous

Published on 13 Apr 2026 - 4:41

score 484 · Answer 4 · 2026-04-12 03:09:26

484

Views

A subset of $\mathbb{R}$ is compact iff it is sequentially compact

Published on 12 Apr 2026 - 3:09

score 285 · Answer 5 · 2015-11-06 23:57:52

285

Views

Why is $C=\{x\in l^2:|x_n|\leq 2^{-n},n=1,2,3,\dots\}$ compact?

Published on 06 Nov 2015 - 23:57

score 271 · Answer 6 · 2026-04-01 18:03:50

271

Views

compactness of the Stone-Čech compactification by ultrafilters

Published on 01 Apr 2026 - 18:03

score 58 · Answer 7 · 2026-04-14 02:59:49

58

Views

Determine why these sets are necessarily open, closed or compact where $f$ is a continuous function

Published on 14 Apr 2026 - 2:59

score 199 · Answer 8 · 2026-03-29 22:33:46

199

Views

Compactness of the closed unit ball of ${\rm Lip}_0(X)$ in the topology of pointwise convergence implies that ${\rm Lip}_0(X)$ is a dual space

Published on 29 Mar 2026 - 22:33

score 1.9k · Answer 9 · 2026-04-12 06:26:21

1.9k

Views

Countably infinite union of compact sets is compact

Published on 12 Apr 2026 - 6:26

score 52 · Answer 10 · 2026-04-11 13:15:59

52

Views

A little question about the proof of compactness implying limit point compactness .

Published on 11 Apr 2026 - 13:15

score 650 · Answer 11 · 2026-04-12 02:51:45

650

Views

Antiderivative as an integral operator from $L^2(0,2\pi)$ to $L^2(0,2\pi)$

Published on 12 Apr 2026 - 2:51

score 107 · Answer 12 · 2026-04-17 08:54:29

107

Views

Must the index for the sets in a finite open subcover $\mathscr{G}$ be arbitrary or can it be a specific number?

Published on 17 Apr 2026 - 8:54

score 676 · Answer 13 · 2026-04-13 00:04:37

676

Views

Locally Compact Metric Space Properties

Published on 13 Apr 2026 - 0:04

score 50 · Answer 14 · 2026-04-09 16:14:54

50

Views

Non-compactness of $(1,3)$ and $[x,\infty)$ only using the definition

Published on 09 Apr 2026 - 16:14

score 1.7k · Answer 15 · 2026-04-11 15:00:17

1.7k

Views

If $A$ is closed and $B$ is compact in $\mathbb R^n$ then $A+B=\{a+b : a \in A \text{ and } b \in B\}$ is closed.

Published on 11 Apr 2026 - 15:00

List Question

Why aren't the rationals a compact subset of $\mathbb{R}$?

Show that a set is weakly compact nonseparable.

Show that $T^$ is weak$^$-to-weak$^*$ continuous

A subset of $\mathbb{R}$ is compact iff it is sequentially compact

Why is $C=\{x\in l^2:|x_n|\leq 2^{-n},n=1,2,3,\dots\}$ compact?

compactness of the Stone-Čech compactification by ultrafilters

Determine why these sets are necessarily open, closed or compact where $f$ is a continuous function

Compactness of the closed unit ball of ${\rm Lip}_0(X)$ in the topology of pointwise convergence implies that ${\rm Lip}_0(X)$ is a dual space

Countably infinite union of compact sets is compact

A little question about the proof of compactness implying limit point compactness .

Antiderivative as an integral operator from $L^2(0,2\pi)$ to $L^2(0,2\pi)$

Must the index for the sets in a finite open subcover $\mathscr{G}$ be arbitrary or can it be a specific number?

Locally Compact Metric Space Properties

Non-compactness of $(1,3)$ and $[x,\infty)$ only using the definition

If $A$ is closed and $B$ is compact in $\mathbb R^n$ then $A+B=\{a+b : a \in A \text{ and } b \in B\}$ is closed.

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