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Prove that there cannot exist any non constant continuous function $f: \mathbb{R} \rightarrow \mathbb{Q}$.

If there exist such an continuous function , it will map interval to a connected subset of $\mathbb{Q}$. I am unable to imagine connected subset of $\mathbb{Q}$ to for proving the above statement. Can anyone help me in this? Thanks.

mea43
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  • Your insight about connected subsets of $\mathbb{Q}$ is good. The only connected subsets of $\mathbb{Q}$ are singletons, sets with one point. It follows that a function with such an image... is a constant function! – hardmath Mar 02 '16 at 12:29

2 Answers2

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A function $\mathbb R\to\mathbb Q$ is in particular also a function $\mathbb R\to\mathbb R$ -- and if you unfold the definition of continuity you will see that the function is still continuous when viewed as $\mathbb R\to\mathbb R$.

Now assume that there is $f(x_1)\ne f(x_2)$ and apply the intermediate value theorem ...

hmakholm left over Monica
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Consider any subset $A$ of $\mathbb{Q}$ which has at least two points, say $a, b$, with $a < b$. Then between $a$ and $b$ there is an irrational number. This tells you that $A$ is disconnected.

Andreas Caranti
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  • Between $1/2$ and $1$ lies for instance $\pi/4$. The sets $U = { x \in \mathbb{Q} : x < \pi/4}$ and $V = { x \in \mathbb{Q} : x > \pi/4}$ are open in $\mathbb{Q}$, and $\mathbb{Q} = U \cup V$, so your set $A$ splits as $A = (A \cap U) \cup (A \cap V)$. (This was a response to a now deleted comment.) – Andreas Caranti Mar 02 '16 at 12:42