Is the following set is compact Consider the set of all $n \times n$ matrices with determinant equal to one in the space of $\mathbb R^{n\times n}$. 
My idea is compact because determinant function is continous ant it is bijective from the given set to $\mathbb R$ and $\mathbb R$ is Hausdorff, so image of compact set  is compact.

In $\mathbb R^m$, with the usual metric topology, a set is compact iff it is closed and bounded. The set you describe is closed (since it is the inverse image of a closed set under a continuous function) but it fails to be bounded if $n>1$ (and the case $n=1$ is trivial, do you see why?). Try to find matrices with arbitrarily large entries, but still having determinant $1$. 
