Continuous bijection at boundary of open set Suppose $f:U \to V$ is a continuous bijection, where $U \subset \mathbb{R}^n$,$V \subset\mathbb{R}^m$ and $U$ is open.  Suppose further that $U \ni x_n \to x \notin U$.  Then $y_n:=f(x_n)$ may not necessarily converge.  I have two questions:


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*Define $d_n:= \text{distance}(y_n,\partial V)$.  Is it true that $d_n \to 0$?

*If $y_n$ does converge, say to $y$, then must we have $y \in \partial V$?
(Edit: initially I forgot to mention the function is bijective.)

Both of them are incorrect. For example, for $n = m=1$, $U= (0,3\pi/2)$ and $f(x) = \cos x$. Then if $x_n \in U$ and $x_n \to 3\pi/2$, we have $y_n \to 0$ and $0$ is an interior point of $V = [-1, 1)$. 
