The problem is as followed:
Does there exist a function $f$ continuous on $[0,1]\times [0,1] = [0,1]^2$ such that $f([0,1]^2) = (0,1)\times (0,1)$?
My attempt: Suppose such continuous function exists. If we regard the function $f$ as a function that maps $[0,1]^2$ to $\mathbb{R}^2$, then $f([0,1]^2)$ is compact in $\mathbb{R}^2$. However, the set $(0,1)\times (0,1)$ is not closed, particularly not compact, in $\mathbb{R}^2$. Thus, such $f$ cannot exist.
What confuses me is that we can also regard $f$ as a function maps $[0,1]^2$ to the metric subspace $(0,1)\times(0,1)$. In this case the set $(0,1)\times(0,1)$ is not compact since it's not a complete metric space. So, again, we conclude that such $f$ cannot exist.
Are these two arguments correct? I've always being confused by the concepts of open, closed, and compact set, where the last one is independent of the ambient space, while the first two are dependent.
Best Answer
The compactness argument is quite valid, $(0,1)^2$ is not closed so not compact and the continuous image of a compact space is compact.
The completeness argument is not correct. The continuous image of a complete metric space (even within the space) need not be complete, e.g. $\arctan(x): \Bbb R \to \Bbb R$ has image $(-\frac{\pi}{2}, \frac{\pi}{2})$. Completeness is a property of the metric, not of the topology and continuous functions need not preserve metric values (isometries do).
There are also more advanced topological properties that the closed square has and the open one does not, but I think compactness is the simplest solution to this problem.