Show that $g(U)$ is not open Let $U\subset \mathbb R^n$, $V\subset \mathbb R^m$ open, $n>m$, $f:U\to V$ an homeomorphism and $g:U\longrightarrow \mathbb R^n$ define by $$g(x)=(f(x),0)=(f_1(x),...,f_m(x),0...,0).$$
How can I show that $g(U)$ is not open ?
The argument in my exercise is: $$g(U)\subset \{y\in\mathbb R^n\mid y_j=0, j=m+1,...,n\}$$ is not open because it contain no ball of radius $r>0$), but I don't understand the argument.
 A: In fact this is a general statement: in a normed vector space $V$ , a proper vector subspace $W$ is never open.
The argument is the following. If $W$ is a proper vector subspace, there exist a vector $x \notin W$. But then if you take an open ball centered at the origin of radius $r > 0$ denoted $B(0,r)$, the vector $y=r \frac{x}{2 \Vert x \Vert} \notin W$ while $y \in B(0,r)$. Hence $W$ cannot be open, as an open is a neighborhood of all its points.
This is what is applied here as $W=\{y\in\mathbb R^n\mid y_j=0, j=m+1,...,n\}$ is a proper vector subspace of $\mathbb R^n$.
A: Let $W$ be a neighbourhood of $g(x)$, then there is some $\delta>0$ such that
$B_\infty(g(x), \delta) \subset W$ (the $\max$ norm ball). Then the point
$p=g(x)+{1 \over 2 } \delta e_{m+1} \in B_\infty(g(x), \delta)$, but clearly
$g(y) \neq p$ for any $y\in U$.
Alternatively, you could use the fact that
a surjective linear map onto a finite dimensional space is open. Then
consider the map $\phi(x) = x_{m+1}$. Since $\phi(g(U)) = \{0\}$, we see that
$g(U)$ is not open (otherwise its image under $\phi$ would be open).
