Let $\pi: \mathbb{R}^2\to \mathbb{R}$ defined by $\pi(x,y)=x$. Let $M$ be a metric space. Prove that $f:[-r,r]\to M$ is continuous if and only if $f\circ \pi: B[0,r]\to M$ is continuous on the closed ball $B[0,r]$ (considering euclidean metric in the interval and euclidean metric in $B[0,r]$).

I only prove the obvious part. In the other part I was proved using Borel-Lebesgue Theorem in a vertical segment of ball. I don't think if it is the idea of exercise because I found in a basic chapter of a book. I want a simpler answer to this fact. I think that is true when continuity is only in the open ball $B(0,r)$, in this case I can't use Borel-Lebesgue Theorem.

The next problem that I have is the following:

Show an example of a set $X\subset \mathbb{R}^2$ such that $\pi(X)=[-1,1]$ and $f:[-1,1]\to\mathbb{R}$ is not continuous but $f\circ \pi: X\to \mathbb{R}$ is continuous.

Obviously I need a subset $X$ such that has a very different topological structure that the closed ball $B[0,1]$. But I don't find it.

  • $\begingroup$ This problem was easier than I've expected. $\endgroup$ – Gaston Burrull May 17 '12 at 4:28

HINT: For the second question, let $f:[-1,1]\to\Bbb R$ be some very simple discontinuous function, and let $X$ be the graph of $f$ as a subset of $\Bbb R^2$.

Added: For the first question, suppose that $f\circ\pi$ is continuous on the open ball $B(0,r)$. Let $$X=\{\langle x,0\rangle\in\Bbb R^2:-r<x<r\}\;,$$ the intersection of $B(0,r)$ with the $x$-axis, and define $$g:(-r,r)\to X:x\mapsto\langle x,0\rangle\;.$$

  1. Is $g$ continuous?
  2. what is the function $(f\circ\pi)\circ g$?
  • $\begingroup$ Wow, really simple example. Thank you very much. Can you help me with the first question pls?, I am so confused. $\endgroup$ – Gaston Burrull May 16 '12 at 23:40
  • $\begingroup$ @Gastón: Not yet, anyway: I’m still thinking about what happens if $f\circ\pi$ is a continuous map from the open ball, so that we don’t have compactness available. $\endgroup$ – Brian M. Scott May 16 '12 at 23:49
  • $\begingroup$ I'm thinking your edited answer. I'll say if I understand all or not. $\endgroup$ – Gaston Burrull May 17 '12 at 0:46
  • $\begingroup$ Great answer $f\pi g$ is evidently $f$ then $f$ is continuous, the same for closed ball. Without using Borel-Lebesgue Theorem. Thank you for help. $\endgroup$ – Gaston Burrull May 17 '12 at 3:55
  • $\begingroup$ Brian I have an alternate answer to my question. It would great if you look for a momment. $\endgroup$ – Gaston Burrull May 17 '12 at 4:16

I was working in an alternate answer to my first question without using Borel-Lebesgue Theorem. Using the metric $d(\left<x,y\right>,\left<z,w\right>)=\max\{|x-z|,|y-w|\}$ in $\mathbb{R}^2$ and the following fact:

Lemma: Let $h:M\times N \to P$ be a function such that $h(x,y)=h(x,z)$ for any $x\in M$, $y,z\in N$. Defining $g: M\to P$ such that $g(x)=h(x,y)$ where $y$ is any element of $N$. Then $g$ is well defined and $g$ is continuous if and only if $h$ is continuous.

Proof: If $\epsilon>0$ note that $d(h(a,b),h(c,d))<\epsilon$ only depends of $a$ and $c$. $\square$

Replacing $h=f\circ\pi$, $g=f$, $M=[-r,r]$ and $N=[-r,r]$, the result follows.

  • 1
    $\begingroup$ Yes, that works too. $\endgroup$ – Brian M. Scott May 17 '12 at 4:23
  • $\begingroup$ Ty for be helpful to me. $\endgroup$ – Gaston Burrull May 17 '12 at 4:23

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