A manifold $M$ of dimension n is a topological space with the following properties:
a) $M$ is Hausdorff
b)$M$ is locally Euclidean of dimension n
c) $M$ has a countable basis of open sets.
Why is the first property necessary? I do not have much experience with Hausdorff spaces, hence I am not able to see the importance of that condition, probably it is something obvious.
Since I come from a physics background, I can understand the importance of the second property. But what is the strong mathematical motivation to study such special subset of a topological space? Also the second property can be alternatively stated as: each point $p\in M$ has a neighborhood $U$ which is homeomorphic to an open subset $U$ of $\mathbb R^n$. Why should it be homeomorphic to an open set and not closed?
Similarly in the third property,why a countable basis of open sets?
Best Answer
A lot of the work on smooth manifolds is to let us use Euclidean analysis to merely locally Euclidean things that come up. Things that aren't Hausdorff are terrible and scary, real intuition busters (at least in my case), so I don't mind at all that we require that. And in fact, with just these 3 requirements (and a smoothness requirement), smooth manifolds can actually be viewed more or less in real space (the Whitney Embedding Theorem). And our real space intuition is pretty good, you know? And it suffices, has really nice properties, provides some pretty rich material, etc.
But there are some people who care a lot about non-Hausdorff manifolds. In fact, secretly, we see them and don't know it. The etale space of the sheaf of continuous real functions over a regular manifold is a manifold, and it's sometimes non-Hausdorff.
Similarly, there are people who care about non-second-countable manifolds. But these are also a bit unfortunate. One of the great things about second countability is that it guarantees that ordinary manifolds are paracompact. A paracompact smooth manifold admits partitions of unity subordinate to a refinement of any cover. Why is this important? (for that matter, what does it really mean?) While it's easy to stitch together continuous functions to make a continuous function (just sort of join the ends together, right?), it's really hard to stitch smooth functions together in general (join the ends together, and perturb it so the first derivatives align, and so the second align, etc.). But this can be done with little fuss with partitions of unity, and thus with little fuss with second-countability.
And if you study smooth manifolds, you'll see that partitions of unity are immediately used for everything.
So it's the way it is because it has these really nice properties, right? Well, why don't we just require manifolds to be paracompact? (Firstly, there is a distinction, but it's 'small.' A manifold with more than countably many disconnected components may be metrizable, and thus paracompact, but obviously won't be second-countable). In this case, the category of paracompact manifolds is closed coproducts, which doesn't hold for second-countable manifolds. In fact, some people do only consider paracompact manifolds.
At the end of the day, Hausdorff and second countability are exactly what let us use the embedding theorem to view manifolds in real space, and that's what's deemed important for people on their first tour through manifolds.