0
$\begingroup$

I want to show that $$f: \mathbb{R}^n \to \mathbb{R}^m, m \leq n, f(x_{1},...,x_{n})=(x_{1},...,x_{m})$$ transforms open sets to open sets. I think it sufficies to show that it transforms open disk into open disk, I'm not sure why and how to start working on that.

$\endgroup$
1
$\begingroup$

If $B=B((x_1,x_2, \ldots, x_n), r)$ is an open ball in $\Bbb R^n$, then show that $f[B]= B((x_1,x_2, \ldots, x_m), r)$ where the last ball is taken in $\Bbb R^m$ of course. The inclusion from left to right is obvious as $f$ decreases distances (we're leaving out $n-m$ coordinates with contributions $\ge 0$ under the square root), and for the other inclusion we can add $x_{m+1}, \ldots, x_n$ to a point in $\Bbb R^m$ and the distance to the centre of $B$ will stay the same as its distance to $(x_1,x_2, \ldots, x_m)$, etc.

$\endgroup$
1
$\begingroup$

Hint: Every norm (hence topology stemming from this norm, i.e. the natural topology) on $\mathbb{R}^d$ is equivalent to any other norm. Take the norm $\Vert x \Vert_{\infty} = \max_{i = 1, \dots, d} |x_i|$, so that your open disks $B(x, r)$ are of the form $A_1 \times A_2 \times \dots \times A_d$ with $A_i = (x_i-r, x_i+r)$.

$\endgroup$
1
  • $\begingroup$ Thanks! I'll try to figure something out :) $\endgroup$ – Nerwena Apr 8 '20 at 11:26
1
$\begingroup$

Since $\mathbb{R^n}$ and $\mathbb{R^m}$ are Banach spaces, and $f$ is a surjective and continuous linear operator, then by the open mapping theorem, $f$ is an open map.

Disclaimer: This theorem is probably "out of reach" from a general topology setting.

$\endgroup$
1
  • $\begingroup$ I didn't have that theorem but I'll check it, it is very helpful! $\endgroup$ – Nerwena Apr 8 '20 at 11:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.