Open Image of a Function Consider the integers $n > m \geq 1$ and let $1 \leq k_1  < \ldots < k_m \leq n$. Define the function $F: \mathbb{R}^n \rightarrow  \mathbb{R}^m$by $F(x_1, \ldots,x_n)=(x_{k_1}, \ldots x_{k_m}) $.
Show that if $U$ is an open subset of $\mathbb{R}^n$, then the image, $F(U)$, is open in $\mathbb{R}^m$.
 A: Pedantic answer: The map $F$ is a continuous linear operator from  the Banach space $\mathbb{R}^n$ onto the Banach space $\mathbb{R}^m$. By the open mapping theorem, $F$ is open. 
Elementary answer: Of course, here, it is trivial as $F$ is a very simple function. Fix $U$ open in $\mathbb{R}^n$ and take $y^0=F(x^0)=F(x_1^0,\ldots,x_n^0)=(x_{k_1}^0,\ldots,x_{k_m}^0)$ in $F(U)$. Since $U$ is open, there exists an open box centered at $x^0$ contained in $U$. Say
$$
B=\prod_{j=1}^n(x_j^0-\epsilon,x_j^0+\epsilon)\subseteq U
$$
for some $\epsilon>0$. Then
$$
y_0\in F(B)=\prod_{j=1}^m(x_{k_j}^0-\epsilon,x_{k_j}^0+\epsilon)\subseteq F(U).
$$
So the open box $F(B)$ is an open neighborhood of $y_0$ in $\mathbb{R}^m$ contained in $F(U)$. This proves that $F(U)$ is open in $\mathbb{R}^m$.
A: If $U$ is open, then for any point $x\in U$ we have some $\epsilon>0$ such that $B_\epsilon(x)\subseteq U$. Thus for any $f(x)\in f[U]$, we have some $\epsilon>0$ such that $f[B_\epsilon(x)]\subseteq f[U]$. Now all you have to do is show that $f[B_\epsilon(x)]=B_\epsilon(f(x))$, .i.e. that the image of the $\epsilon$-ball around $x$ in $\mathbb R^n$ is the $\epsilon$-ball around $f(x)$ in $\mathbb R^m$. This is a straightforward exercise using the definition of an open ball.
