Question

$f(x) = \begin{cases} 0 & x\in( \mathbb{Q}\cap [0,1])^c \\ \frac{1}{q} & x=\frac{p}{q} \in \mathbb{Q} \cap [0,1], (p,q)=1 \end{cases}.$

score 138 · Answer 1 · 2019-02-16 13:33:46

138

Views

$f(x) = \begin{cases} 0 & x\in( \mathbb{Q}\cap [0,1])^c \\ \frac{1}{q} & x=\frac{p}{q} \in \mathbb{Q} \cap [0,1], (p,q)=1 \end{cases}.$

Published on 16 Feb 2019 - 13:33

score 72 · Answer 2 · 2026-03-25 17:38:56

72

Views

If a function $f: A \to \mathbb R$ is bounded and continuous on $A$, does its integral on said interval converge uniformly?

Published on 25 Mar 2026 - 17:38

score 32 · Answer 3 · 2019-02-16 17:45:13

32

Views

Verify that $f(x)=\frac{1}{x}-\frac{1}{x_0}$ is continuous for every $x_0\neq 0$.

Published on 16 Feb 2019 - 17:45

score 150 · Answer 4 · 2026-03-27 21:20:30

150

Views

Show that $K∩L$ is homeomorphic to $A$, where $K ∩ L = \{\{(a,0),(a,1)\} : a \in A\}$ is a subset of equivalence classes in $[0,1] \times \{0,1\}$.

Published on 27 Mar 2026 - 21:20

score 116 · Answer 5 · 2019-02-17 06:13:03

116

Views

For what values of $a$ and $b$ is the function $\frac{x^ay^b}{x^2+y^2}$ continuous at $(0,0)$?

Published on 17 Feb 2019 - 6:13

score 132 · Answer 6 · 2019-02-17 07:34:58

132

Views

I am stuck somewhere in this analysis problem

Published on 17 Feb 2019 - 7:34

score 177 · Answer 7 · 2026-04-08 16:19:31

177

Views

Preimage of continuous one-to-one function on connected domain is not continuous.

Published on 08 Apr 2026 - 16:19

score 55 · Answer 8 · 2026-03-25 12:55:01

55

Views

Prove that this group action is continuous from $S\times\Bbb{R}^2\to\Bbb{R}^2$

Published on 25 Mar 2026 - 12:55

score 106 · Answer 9 · 2026-03-25 04:44:19

106

Views

What does continuity of the determinant say about the value of a submatrix on neighborhoods of $M_{m\times n}(\mathbb{R})$

Published on 25 Mar 2026 - 4:44

score 739 · Answer 10 · 2026-03-25 19:01:57

739

Views

Why does weak-$L^2$ convergence not imply pointwise convergence for continuous functions?

Published on 25 Mar 2026 - 19:01

score 71 · Answer 11 · 2019-02-17 22:58:23

71

Views

Is a non-negative function which satisfies the parallelogram identity continuous?

Published on 17 Feb 2019 - 22:58

score 171 · Answer 12 · 2026-04-07 22:44:59

171

Views

convert iterated function to continuous function

Published on 07 Apr 2026 - 22:44

score 47 · Answer 13 · 2026-04-08 20:09:04

47

Views

Fixed value property of $[0,1]$

Published on 08 Apr 2026 - 20:09

score 1k · Answer 14 · 2026-03-27 13:46:04

1k

Views

Integrating continuous function of two variables by one of the variables, do I get continuous function?

Published on 27 Mar 2026 - 13:46

score 96 · Answer 15 · 2019-02-18 09:09:31

96

Views

Proof verification there exist at most one $c\in[0,1]$ such that $f(c)=c$.

Published on 18 Feb 2019 - 9:09

List Question

$f(x) = \begin{cases} 0 & x\in( \mathbb{Q}\cap [0,1])^c \\ \frac{1}{q} & x=\frac{p}{q} \in \mathbb{Q} \cap [0,1], (p,q)=1 \end{cases}.$

If a function $f: A \to \mathbb R$ is bounded and continuous on $A$, does its integral on said interval converge uniformly?

Verify that $f(x)=\frac{1}{x}-\frac{1}{x_0}$ is continuous for every $x_0\neq 0$.

Show that $K∩L$ is homeomorphic to $A$, where $K ∩ L = \{\{(a,0),(a,1)\} : a \in A\}$ is a subset of equivalence classes in $[0,1] \times \{0,1\}$.

For what values of $a$ and $b$ is the function $\frac{x^ay^b}{x^2+y^2}$ continuous at $(0,0)$?

I am stuck somewhere in this analysis problem

Preimage of continuous one-to-one function on connected domain is not continuous.

Prove that this group action is continuous from $S\times\Bbb{R}^2\to\Bbb{R}^2$

What does continuity of the determinant say about the value of a submatrix on neighborhoods of $M_{m\times n}(\mathbb{R})$

Why does weak-$L^2$ convergence not imply pointwise convergence for continuous functions?

Is a non-negative function which satisfies the parallelogram identity continuous?

convert iterated function to continuous function

Fixed value property of $[0,1]$

Integrating continuous function of two variables by one of the variables, do I get continuous function?

Proof verification there exist at most one $c\in[0,1]$ such that $f(c)=c$.

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