Let $A \subset R^k$ be an open set and $f:A \to R^n$ be a continuous injection $(k < n)$.
The task is to prove that:
$f:A \to f(A)$ is not a homeomorphism if and only if there exists a sequence $\{ x_k \} \subset A$ that converges to some point on the boundary of $A$ or to $\infty$ so that the $ a = \lim\limits_{k \to \infty} f(x_k)$ exists and belongs to $f(A)$.
First it can be noticed that $f: A \to f(A)$ is a bijection. The second thing is that $f$ is continuous, since it was given. So' the only missing thing here for $f$ to be a homomorphism is that $f^{-1}$ needs to be continuous.
In the first direction, I assume that such a sequence exists. If $x_k$ converges to a point on the boundary or to $\infty$ then I can take the sequence $\{ f(x_k) \}$ which converges to $a$, however $\lim\limits_{k \to \infty} f(x_k) = a$ and $f^{-1}(a) \ne \lim\limits_{k \to \infty} f^{-1}(f(x_k))$, since it is given that $a$ belong to $f(A)$ so there is some $y \in A$ that satisfies $f(y) = a$.
So I thing I managed to prove here the first direction.
However, I don't really know how to start in the second direction.
Does anyone here has any suggestion?
Help would be appreciated.
Well, if $f$ is not a homeomorphism, then $f^{-1}$ is not continuous, thus there exists a convergent sequence $y_n=f(x_n)$ of elements of $f(A)$ such that $f^{-1}(y_n)=x_n$ does not converge to an element of $A$.