# Show that projection transforms open sets to open sets

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.

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.
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 $$A_i = (x_i-r, x_i+r)$$.
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.