Why is $[0,1]$ not homeomorphic to $[0,1]^2$?

Question

Why is $[0,1]$ not homeomorphic to $[0,1]^2$? It seems that the easiest way to show this is to find some inconsistency between the open set structures of the two. It is clear that the two share the same cardinality. Both are compact. Both are normal since they are metric spaces. However, where to find the open set structure that is not shared by the two? Any hint, please?

Answer 1

If $A$ is omeomorphic to $B$ through $f$, then $A \setminus \lbrace a \rbrace$ is omeomorphic to $B \setminus \lbrace f \left( a \right) \rbrace$ through $f$. Then pick $a= \frac{1}{2}$. $\left[ 0,1\right] \setminus \lbrace \frac{1}{2} \rbrace$ is disconnected, while $\left[ 0,1\right] \times \left[ 0,1\right] \setminus \lbrace p \rbrace$ is connected for any $p \in \left[ 0,1\right] \times \left[ 0,1\right]$.

Answer 2

If you remove an inner point from $[0,1]$, the resulting space is disconnected. This is clearly false if you remove any point from $[0,1]\times [0,1]$.

This is a rather rough approach. A much more general one follows from Dimension Theory, or from the Invariance-of-domain Theorem. It is stated for open subsets, but I think it is pretty easy to deduce your statement from the statement that $(0,1)$ is not homeomorphic to $(0,1) \times (0,1)$.

Answer 3

As comments and other questions point out, this case is easily solved by a connectedness argument.

It may be worth to point out that the general proof that $\mathbb R^n$ is not homeomorphic to $\mathbb R^m$ for $m\neq n$ is not trivial.

There is a nice important result, called the invariance of domain, (see http://en.wikipedia.org/wiki/Invariance_of_domain) that states that if a map $f:U\subset \mathbb R^n\to \mathbb R^n$ is continuous and injective, then it is open. This implies that there is no continuous injective map from $\mathbb R^{n+k}$ to $\mathbb R^n$.