# Continuity of a function

1. Sep 10, 2007

### ELESSAR TELKONT

My problem is this. Let $$f:\mathbb{R}^{2}\longrightarrow \mathbb{R}^{2}$$ be a continuous function that satifies that $$\forall q\in\mathbb{Q}\times\mathbb{Q}$$ we have $$f(q)=q$$. Proof that $$\forall x\in\mathbb{R}^{2}$$ we have $$f(x)=x$$.

I have worked out that because it is continuous, $$f$$ satisfies that
$$\forall \epsilon>0\exists\delta>0\mid \forall x\in B_{\delta}(a)\longleftrightarrow f(x)\in B_{\epsilon}(f(a))$$

and then $$\forall q\in\mathbb{Q}\times\mathbb{Q}$$ we have
$$\forall \epsilon>0\exists\delta>0\mid \forall x\in B_{\delta}(q)\longleftrightarrow f(x)\in B_{\epsilon}(q)$$

therefore we have to proof that $$\forall x'\in\mathbb{R}^{2}$$ we have
$$\forall \epsilon>0\exists\delta>0\mid \forall x\in B_{\delta}(x')\longleftrightarrow f(x)\in B_{\epsilon}(x')$$.

It's obvious that every element of $$\mathbb{R}^{2}$$ could be approximated by some element of $$\mathbb{Q}\times\mathbb{Q}$$ or sequence in this. But, how I can link this in an expression to get what I have to proof?

2. Sep 10, 2007

### TimNguyen

I'm not so sure about this but do we not know how the function maps irrational numbers, such as sqrt(2)?

3. Sep 10, 2007

### CompuChip

TimNguyen, in fact we know (they are mapped to themselves, as f is the identity map) but this is exactly what Elessar Telkont wants to show.

Indeed you got the idea right: any real number can be approximated by a sequence of rational numbers (and therefore, pairs of reals can be approximated by pairs of rationals).
What I would do is: Try to make this process of approximation precise (describe it in terms of epsilon-delta). Now assume what you want to prove is not true, then this should give a contradiction with the continuity (which you have also written out in epsilon-delta).

I will take a look and post it more precisely later on (first, you give it a try yourself)

4. Sep 10, 2007

### CompuChip

Hmm, it was much easier.

It is a familiar fact (or otherwise you should be able to easily prove it from the definition) that for continuous functions f, it holds that $$\lim_{n \to \infty} f(x_n) = f(\lim_{n \to \infty} x_n)$$ for a sequence $$(x_n)_{n \in \mathbb{N}}$$.
So describe a real pair x as the (coordinate-wise) limit of a sequence $x_n$ of rational pairs, then
$$f(x) = f(\lim_{n \to \infty} x_n) \stackrel{*}{=} \lim_{n \to \infty} f(x_n) = \lim_{n \to \infty} x_n = x,$$
where the identity marked with a star holds by the continuity of f -- QED.

For completeness, let me prove the claim about the limits (it's a nice exercise in epsilon-delta proofs, so you might want to try it yourself first):
Let $\epsilon > 0$. Since f is continuous, there is some $\delta$ such that $|| x - x_n || < \delta$ implies that $|| f(x_n) - f(x) || < \epsilon$.
Now $x_n$ converging to x means that for this $\delta$ I can find an $N$ such that $|| x_n - x || < \delta$ as long as $n > N$.
So, through the $\delta$ from the definition of continuity, I have found an $N$ for my $\epsilon$ such that $n > N$ implies $|| f(x_n) - f(x) || < \epsilon$, in other words,
$$\lim_{n \to \infty} f(x_n) = f(x) = f( \lim_{n \to \infty} x )$$.

Last edited: Sep 10, 2007
5. Sep 10, 2007

### TimNguyen

Sorry about that. My math is extremely rusty since I started graduate school in physics.