Linked Questions

7
votes
3answers
3k views

Continuous function that take irrationals to rationals and vice-versa. [duplicate]

Can someone help me? How can I prove that there isn't an everywhere continuous function $f:\mathbb R \rightarrow \mathbb R$ that transforms every rational into an irrational and vice-versa?
3
votes
4answers
334 views

Continuous function $f$ on $\mathbb R$ such that $f(\mathbb Q)\subseteq \mathbb R-\mathbb Q$ and $f(\mathbb R-\mathbb Q)\subseteq \mathbb Q$? [duplicate]

Possible Duplicate: No continuous function that switches $\mathbb{Q}$ and the irrationals Is there a continuous function $f\colon\mathbb R\to \mathbb R$ such that $f(\mathbb Q)\subseteq \mathbb R-...
5
votes
2answers
293 views

$f(\mathbb{R}\setminus \mathbb{Q}) \subseteq \mathbb{Q}$ and $f(\mathbb{Q}) \subseteq \mathbb{R}\setminus \mathbb{Q}$ imply that $f$ is not continuous [duplicate]

Possible Duplicate: No continuous function that switches $\mathbb{Q}$ and the irrationals Let $f: \mathbb{R} \to \mathbb{R}$ be function satisfying the two conditions: $f(\mathbb{R}\setminus \...
1
vote
3answers
544 views

Proving that there is no continuous function $f:\Bbb R\to\Bbb R$ satisfying $f(\Bbb Q)\subset\Bbb R-\Bbb Q$ and $f(\Bbb R-\Bbb Q) \subset\Bbb Q$. [duplicate]

How can I prove that there is no continuous function $f:\mathbb{R}\to \mathbb{R}$ satisfying $f(\mathbb{Q}) \subset \mathbb{R}\backslash \mathbb{Q}$ and $f(\mathbb{R}\backslash \mathbb{Q} ) \subset \...
2
votes
1answer
401 views

Nowhere Continuous Function [duplicate]

I was reading Dirichlet and Thomae's functions and got interested to know about functions which are continuous nowhere. Since these have a lot to do with rationals and irrationals, the next question ...
2
votes
2answers
115 views

$f(x)\in \mathbb Q,$ if $x\not\in \mathbb Q$ and $f(x)\not\in \mathbb Q$, if $x\in \mathbb Q$. [duplicate]

$f(x)\in \mathbb Q,$ if $x\not\in \mathbb Q$ and $f(x)\not\in \mathbb Q$, if $x\in \mathbb Q$. Can $f$ be continuous? I have tried using the sequential definition of continuity on rational and ...
4
votes
1answer
90 views

There exists a continuous function $f$ such that $f(\Bbb Q) \subseteq \Bbb R\setminus\Bbb Q$ and $f(\Bbb R\setminus\Bbb Q)\subseteq\Bbb Q$ [duplicate]

True or false: There exists a continuous function $f: \Bbb R \to \Bbb R$ such that $f(\Bbb Q) \subseteq {\Bbb R}\setminus {\Bbb Q}$ and $f({\Bbb R}\setminus {\Bbb Q}) \subseteq {\Bbb Q}$. My attempt:...
0
votes
2answers
69 views

Continuous Function taking rationals to irrationals and vice versa [duplicate]

Note Do not close this question like It was done earlier. Question is different so I am asking a new question. l am not supposed to use connectedness here Mine is a basic real analysis course. ...
3
votes
1answer
102 views

Does this type of functions exist [duplicate]

Does there exist a cont function $f$ : $\mathbb{R}$ $\rightarrow$$\mathbb{R}$ which takes irrational values at rational points and rational values at irrational points?
2
votes
1answer
110 views

Existence of Continuous function [duplicate]

Does there exist a continuous function $f : \Bbb R \to \Bbb R$ which takes irrational values at rational points and rational values at irrational points?
0
votes
1answer
39 views

A single-variable continuous function that is irrational if and only if its argument is rational [duplicate]

I wonder if it is possible to construct a continuous $f : \mathbb R \to \mathbb R$ such that for each $x \in \mathbb R$, $f(x)$ is irrational if and only if $x$ is rational? my attempt Sadly, I ...
0
votes
1answer
69 views

Can such a function be continuous? [duplicate]

Let $f$ be a function from $\Bbb R$ to $\Bbb R$ such that $f(x)$ is rational when $x$ is irrational, and $f(x)$ is irrational when $x$ is rational. Can $f$ be continuous? Thanks for your help.
0
votes
1answer
60 views

Existence of a continuos function in $[0, 1]$ [duplicate]

Can exist a continuous function $f:[0, 1]\rightarrow \mathbb{R}$ so that $f(x)\in\mathbb{Q}$ if $x\in\mathbb{I\cap [0, 1]}$ and $f(x)\in\mathbb{I}$ if $x\in\mathbb{Q\cap [0, 1]}$? Why yes? why not?
-1
votes
1answer
30 views

does there exist example of this paticular type of function? [duplicate]

consider $f:[a,b]\to \mathbb{R}$ such that $f(x) \in \mathbb{Q}$ when $x \in \mathbb{R} \setminus \mathbb{Q} \cap [a,b]$ and $f(x) \in \mathbb{R} \setminus \mathbb{Q}$ when $x \in \mathbb{Q} \cap [a,b]...
20
votes
7answers
43k views

Show that a continuous function has a fixed point

Question: Let $a, b \in \mathbb{R}$ with $a < b$ and let $f: [a,b] \rightarrow [a,b]$ continuous. Show: $f$ has a fixed point, that is, there is an $x \in [a,b]$ with $f(x)=x$. I suppose this has ...

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