Functions mapping an open interval to a closed one

Prove that there exist non constant functions $f:\mathbb{R} \rightarrow \mathbb{R}$ that map any open interval onto a closed one.

Solution   $\newcommand{card}{{\rm card}}$

Let $\mathcal{A}=\{ [x] : x \in \mathbb{R}\}$ be the set of the equivelance classes. First we are proving a lemma:

Lemma: $\card \mathcal{A}=\card \mathbb{R}$

Proof:

It is pretty well known that the dimension of $\mathbb{R}$ as vector space over $\mathbb{Q}$ is $\aleph$, where $\aleph=\card \mathbb{R}$. If we choose a basis $\mathcal{B}$ of $\mathbb{R}$ over $\mathbb{Q}$ which contains $1$ (it is possible because any linearly independent set can be extended to a basis and $\{1\}$ is such set. ) If two di fferent numbers, say $x$ and $y$ from $\mathcal{B}$ would be in the same
equivalence class, then we have $x-y =q \cdot 1 , \; q \in \mathbb{Q}$ so $1$, $x$ and $y$ are linear dependent ,a contradiction. Therefore any two different elements of $\mathcal{B}$ are in different equivelance classes and we can define an one to one mapping $\phi: \mathcal{B} \rightarrow \mathcal{A}, \quad \phi(b)=[b] \; \forall b \in \mathcal{B}$. This yields $\card \mathcal{B} \leq \card \mathcal{A}$. However, from the early lemma we stated at the beginning of this proof , we know that $\card \mathcal{B} = \aleph$ and also that $\card \mathcal{A} \leq \aleph$. The result follows.

We will prove more than this, that is there exist such functions which map any open interval into the same closed interval. Take$a, b \in \mathbb{R}$ such that $a<b$. Then $\card[a, b] = \aleph = \card \mathcal{A}$ so we can find a bijective function $g: \mathcal{A} \rightarrow [a, b]$ . We define the function $f: \mathbb{R} \rightarrow [a, b]$ as $f(x)=g([x])$. We will prove that the image or any interval $(s, t)$ such that $s<t \quad s, t \in \mathbb{R}$ through $f$ is exactly $[a, b]$.

First of all we notice that $[x]= x + \mathbb{Q}=\{x + q , \; q \in \mathbb{Q} \}$. Because $\mathbb{Q}$ is dense in $\mathbb{R}$ any of its translations is also dense in $\mathbb{R}$  meaning that any equivalence class from $\mathcal{A}$ has at least one element in common with any open interval , $(s, t)$ being such an interval. The result follows.