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I have been attempting to do the following question by contradiction. However, I just got stuck at where to use the given continuity condition. It would be really appreciated if you can possibly give a further hint to solving this problem.

Here it goes

Let $f(x):\mathbb{R} \rightarrow \mathbb{Q}$ is a continuous function. Prove that $f$ is a constant.

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    Look at [t$h$is]($h$ttp://math.stackexchange.com/q/55638/8271)2019-02-10

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The basic idea behind this problem is the following statement:

Theorem: The continuous image of a connected space is connected.

Now, $\mathbb{R}$ is connected--that is just a fact. Now, $\mathbb{Q}$ is not connected (in fact, it's totally disconnected) since given any irrational number $\xi$ one has that $\mathbb{Q}=[(-\infty,\xi)\cap\mathbb{Q}]\cup[(\xi,\infty)\cap\mathbb{Q}]$ is a disconnection.

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    @VietHoangQuoc In the last comment, the problem you tell us is best solved using $\varepsilon-\delta$ definition is more appropriate.2012-05-05