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 $f(\mathbb{R}\backslash \mathbb{Q} ) \subset \mathbb{Q}$?
1369
Is there a way to prove the following result using connectedness?  Result:  Let $J=\mathbb{R} \setminus \mathbb{Q}$ denote the set of irrational numbers. There is no continuous map $f: \mathbb{R} \rightarrow \mathbb{R}$ such that $f(\mathbb{Q}) \subseteq J$ and $f(J) \subseteq \mathbb{Q}$.
21174
Let $C(\mathbb{R})=\{ f:\mathbb{R}\to \mathbb{R} \colon \ f \text{ continuous}\}\subseteq \mathbb{R}^{\mathbb{R}} $. How to prove formally that $C(\mathbb{R}) \notin \mathcal{B}(\mathbb{R}^{\mathbb{R}})$ (the product sigma algebra)?  The classic proof said that   $A \in \mathcal{B}(\mathbb{R}^{\mathbb{R}})$ iff exists $J\subseteq \mathbb{R}$ with $|J|\leq \aleph_0$ and $B\in \mathcal{B}(\mathbb{R^J})$ such that $A=B \times \mathbb{R}^{\mathbb{R}\setminus J}=\{f \in \mathbb{R}^\mathbb{R} \colon  \ (f(j) \colon \ t \in J) \in B\}$. In simpler terms,  $A$ only can have countable restrictions but the set of continuous functions have uncountable restrictions since every function on $C(\mathbb{R})$ needs to be continuous in every point $x \in \mathbb{R}$.  The last part of this proof always didn't satisfied me at all, and I don't really know how to do this proof formally.   Any help or references will be appreciated.
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I was thinking, inspired by mathlinks, precisely from , if there exists a continuous real function $f:\mathbb R\to\mathbb R$ such that $$f(f(x))=e^x.$$ However I have not still been able to come up with an answer. I would like to share this problem with you.   I'm not aware of its level, though i wouldn't classify it as homework, so I'm not giving it the homework tag. If anybody feels that this problem is in reality easy or looks like an homework, please feel free to add that tag.    EDIT for those interested in the complex case, I've found this on MathOverflow. It's the first answer. Wow!
40644
How does one compute the cardinality of the set of functions $f:\mathbb{R} \to \mathbb{R}$ (not necessarily continuous)?
3532
Does there exist a continuous function $\: f : \mathbf{R} \to \mathbf{R} \:$ such that for   all  functions $\: g : \mathbf{R} \to \mathbf{R} \:$, for all real numbers $x$,  there exists a real number $y$ such that $\: x &lt; y \:$ and $\: g(y) &lt; f(y) \:$?
51565
Find all continuous functions $f:\mathbb R^{+}\to\mathbb R^{+}$ such that $f(x)^2=f(x^2)$ for all positive reals $x$.  If $f$ is a constant function, then $f(x)=1$. If $f$ is non-constant, I'm suspecting the only solutions are of the form $f(x)=x^k$, where $k$ is a constant, but I have no idea how to prove this. Thanks in advance.
58164
I'm trying to find a surjective function $f:\mathbb{N}\to \mathbb{Q}$;  I know that at least one such function must exist since $\mathbb{Q}$ is countable, but I haven't been able to find one.  Can someone show me one such function?  Best regards,  lorenzo
23132
Assume $f$ is a function over $\mathbb{R}$ satisfying $f(x+y) = f(x)+f(y)$ for all $x,y \in \mathbb{R}$. Show that if $f$ is continuous, then $f(x) = cx$ for all $x \in \mathbb{R}.$   I find it hard to use the definition of continuity to derive anything from the equation. I think we could write the definition of the limit in terms to get $$\forall \epsilon,\exists \delta \quad 0&lt;|x-a|&lt;\delta \quad \implies \quad |f(x)-L| &lt; \epsilon.$$ I am not sure where to go from here.
92909
What are some nice examples of continuous functions $f : \mathbb{R} \to \mathbb{R}$ where the images of open sets are not open, and the images of closed sets are not closed?
59414
How would you show that there is no continuous injective map  $f:\mathbb{R}^3 \to \mathbb{R}$?  I tried the approach where if $a \in \mathbb{R}$ then $\mathbb{R} \backslash \{a\} $ is not clearly connected so there can't exist a continuous function otherwise $\mathbb{R} \backslash \{a\} $ would be connected?
115125
Does there exist a continuous function $f: \mathbb{R} \to \mathbb{R}$ that takes each value in $\mathbb{R}$ exactly two times?
2291
Question is to prove :  $f : S^1 \to \mathbb R$ is continuous then $f(x)=f(-x)$ for some $x\in S^1$  I guess it would be helpful to use intermediate value theorem  Assuming $f(x)\neq f(-x)$ then given any $p\in (f(-x),f(x))$ (assuming $f(-x)&lt;f(x)$) there exists $y\in (-x, x)$ such that $f(y)=p$  I am not very sure of how to use this..   It would be helpful if someone can give some hint which would help me to solve this..  Thank you.
39321
Let $f:[0,1]\to \mathbb{R}$ is continuous, such that $f(0)=f(1)$. For each $n\in\mathbb{N}$, prove that $\exists x\in [0,1]$ such that $x+\dfrac{1}{n}\in[0,1]$ and $f\left(x+\dfrac{1}{n}\right)=f(x)$.   My approach: The interval $[0,1]$ is connected, then if $f$ is continuous, we have $f([0,1])\subset\mathbb{R}$ is a interval. And I suppose that, I can define a closed path such that $f(0)=f(1)$. Any hint pls,
87288
The question is: Prove that if $f:\mathbb{R}\to\mathbb{R}$ is continuous and $f(x) = 0$ whenever $x$ is rational, then $f(x) = 0$ for all $x\in\mathbb{R}$.  My proof:  Let $ x\in\mathbb{R} $. If $ x\in\mathbb{Q}$, then $ f(x)=0 $.  Otherwise $ x\in\mathbb{R\backslash Q} $. Then there exists a sequence of rational numbers $ (q_n)_{n\in\mathbb{N}}$ (for example $q_n= \frac{\lfloor{nx\rfloor}}{n} ,\forall n\in\mathbb{N}$) such that $q_n\to x$ as $n\to \infty$. By the properties of $f$, we have $f(q_n)=0$ for all $n\in\mathbb{N}$. So $f(q_n)\to 0$ as $n\to \infty$. Since $f$ is continuous, $q_n\to x$ implies that $f(q_n)\to f(x)=0$. So $f(x) = 0$ for all $x\in\mathbb{R}$.   Thank you for any help.
16889
So the question asks to find a topology on $\mathbb{R}$ so that all the polynomials are continuous functions from $\mathbb{R}$ to $\mathbb{R}$ but $\cos(x)$ is not continuous as a function from $\mathbb{R}$ to $\mathbb{R}$.   I looked at another similar question in stack exchange:  I think this is a special case of my question since it requires all other functions to not be continuous.   Other than that I thought maybe the solution could be based on how cosine is a periodic function unlike the polynomials, so maybe I should take out some infinite sets from the topology but that still wouldn't be enough since the topology would be closed under unions. So I am not even sure if there exists such topology.
36082
I have the following question, more general that   Let $U\subset\mathbb{R}^n$ be a domain. If $f:U\longrightarrow\mathbb{R}^m$  ($m&lt; n$) is a continuous function, how to prove that $f$ is not injective.  Any hint would be appreciated.
13998
I have a question about subsets $$ A \subseteq \mathbb R $$ for which there exists a function $$f : \mathbb R \to \mathbb R$$ such that the set of continuity points of $f$ is $A$. Can I characterize this kind of sets? In a topological,measurable or in some way? For example, does there exist a function continuous on $\mathbb Q$ and discontinuous on the irrationals?