Is Every Point Mapped to Itself in a Continuous Function on ℝ²?

[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?

 

[TimNguyen]

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

 

[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)

 

[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.

[edit]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 ).

 

[TimNguyen]

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)

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