• Oct 16th 2012, 02:22 AM
vercammen
Let
$A = \{(x,y) \in \mathbb{R}^2 | (x+1)^2 + y^2 = 1\}$,\\

$B = \{(x,y) \in \mathbb{R}^2 | (x-1)^2 + y^2 = 1\}$, and\\
$C = \{(x,y) \in \mathbb{R}^2 | x^2 + y^2 = 1\}$.\\

Define $N = A \cup B$
and let $N$ and $C$ be given subspaces of ${\mathbb{R}}^2$, where ${\mathbb{R}}^2$ is given the standard topology. Show that $N$ is not homeomorphic to $C$.

PICTURE ATTACHED!
• Oct 16th 2012, 11:07 PM
johnsomeone
The way to do kind of problem (show that two spaces aren't homeomorphic) is usually to find a topological property (i.e. something that must be the same for homeomorphic spaces) that's not shared by N and C.

Connectedness, path connectedness and compactness are the first ones you learn about, but there are gobs more. There's an entire subject called algebraic topology dedicated to associating abstract number systems (all manner of groups and rings) to topological spaces in a way that if the associated abstract objects are different, then the spaces can't be homeomorphic. Existence of maps to or from some other space is also a topological property. Any local property (every point of X has a neighborhood such that...) are topological properties.

An obvious one here is that C is a manifold, but N is not, because of where those two circles come togther in a "cross" on N.

C is a manifold because it locally "looks like" $\displaystyle \mathbb{R}^1$ (every pt in C is in an open set that's homeo to $\displaystyle \mathbb{R}^1.$)

However, even though it's visually obvious that N isn't locally homeomorphic to $\displaystyle \mathbb{R}^1$, "obvious" isn't a proof.

The most elementary way to solve this is to consider what's different between them if you remove a point. Obviously, for N, you want to remove that intersection point. For C all points are the same, so you get the same thing topologically regardless of which point you remove. Suppose N and C are homeomorphic via f. Then so are N-{point} and C-{f(point)}.

So, among connectedness, path connectedness and compactness, what seems
different between N-{point} and C-{f(point)}?

Can you think of any others, besides the above and my manifold example?
• Oct 16th 2012, 11:36 PM
vercammen