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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.

Alex Youcis
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Viet Hoang Quoc
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1 Answers1

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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.

Pedro
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Alex Youcis
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  • Great solution, however, the question requires the answer to use the definition of the continuity. Thereby, I just do not know where to start. – Viet Hoang Quoc May 05 '12 at 00:39
  • @VietHoangQuoc Well, which definition of continuity are you using? – Alex Youcis May 05 '12 at 00:41
  • The formal one as follows

    For all $ \epsilon >0$, there exists $ \delta >0$ such that for all $ x \in \mathbb{R}$ satisfying $ |x-x_0| < \delta$, one has $ |f(x)-f(x_0)| < \epsilon $.

     – Viet Hoang Quoc May 05 '12 at 00:43
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    @VietHoangQuoc Ditch that definition here. Use the following definition: A function $f: X \rightarrow Y$ between two topological spaces is continuous if for every open set $U$ in $Y$, the preimage of $U$ under $f$ is open in $X$. –  May 05 '12 at 00:46
  • @VietHoangQuoc As Benjamin points out, that definition is often inadequate when dealing with topological problems. Is there any chance you can use the the definition Benjamin pointed out? There is, philosophically, no difference between the two definitions. – Alex Youcis May 05 '12 at 00:48
  • @Benjamin Lim : This is the first course in Analysis so we have not learnt anything such as Topological spaces and other things, just simply the definition and way to go. Thanks. – Viet Hoang Quoc May 05 '12 at 00:50
  • @VietHoangQuoc What if it was phrased in this regards. Every open ball $B_\delta(x)\subseteq\mathbb{Q}$ has preimage equal to the union of balls? Is that closer to something you may be able to work with? – Alex Youcis May 05 '12 at 00:51
  • To my understanding, if we can pick a rational point $x_1$ satisfying $|x_1-x_0|< \delta$ such that $|f(x_1)-f(x_0)| \ge \epsilon$ then this will give a contradiction. – Viet Hoang Quoc May 05 '12 at 00:52
  • @VietHoangQuoc Because now we are dealing with the usual topology on $\Bbb{R}$ and the usual subspace topology on $\Bbb{Q}$, the definitions of continuity using open sets and epsilon - deltas *are not different from each other*. –  May 05 '12 at 00:55
  • Thanks all the the great effort made, let's me just recap what we have had so far: $f$ is continuous iff for all $\epsilon >0$, $ x \in B_{\delta}(x_0)$ then $f(x) \in B_{\epsilon}(f(x_0))$ – Viet Hoang Quoc May 05 '12 at 00:57
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    Here is another similar type of question proving the function is constant using the continuity definition,

    Let $f: [-1,1] \to \mathbb{R}$ be a function with $f(x)=f(x^2)$ for all $ x \in (-1,1)$. Suppose that $f$ is continuous at 0. Prove that $f$ is a constant.

     – Viet Hoang Quoc May 05 '12 at 01:03
  • @VietHoangQuoc In the last comment, the problem you tell us is best solved using $\varepsilon-\delta$ definition is more appropriate. –  May 05 '12 at 19:25