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Please help! Homeomorphism in R^2. Standard Topology.

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!

Re: Please help! Homeomorphism in R^2. Standard Topology.

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?

Re: Please help! Homeomorphism in R^2. Standard Topology.

Does this work?

N is the union of the two circles of unit radius with centres at (-1,0) and (1,0) respectively.Both these circles have only one point in their intersection, which is the point (0,0). Clearly, N\{(0,0)} is a disconnected set, as it is the union of two disjoint open sets A\{(0,0)} and B\{(0,0)|.

If N were homeomorphic to C via some homeomorphism map f: N → C, then N\{(0,0)} would be homeomorphic to C\{f(0,0)} via the restriction of the same map f. But this is not possible, as C\{f(0,0)} is path connected, and hence connected(it is just the unit-radius circle centered at origin. If we remove an point from it, we can still get a path between any two points, by just avoiding that point in our path), but N\{(0,0)} is disconnected(as we have seen above).

[We have used the fact that a homeomorphism maps connected spaces to connected spaces. Here, since N\{(0,0)} is not connected, it can't be homeomorphic to the connected space C\{f(0,0)}].

contradiction

Re: Please help! Homeomorphism in R^2. Standard Topology.

That exactly it.

You should ideally stick with using just one or the other of path connected versus connected, but actually both will work here.