Let's talk topology

Ben · 06-16-2009

Edit: Overlapped.

OK folks, after overnight reflection, I concede Qfwfq's point: neighbourhoods need not be open. The reasons for my conversion will become clear in due course.

So anyway, we now enter the heart of topology. First this;

As always, suppose $X$ is a topological space. A family $\mathcal {A}$ of subsets of $X$ is called a cover for $X$ if $X$ coincides with the union of all $A \in \mathcal{A}$ . Clearly any space has at least one cover of some sort. And if the number of $A \in \mathcal{A}$ is finite (it may not be), then this is called a finite cover.

If each $A \in \mathcal{A}$ is open, it's called an open cover. And if there is a sub-family of $\mathcal{A}$ that also covers $X$ , it is called a sub-cover, naturally enough.

Now, if a finite cover admits of a sub-cover, then this must always be finite. If a cover is not finite, then any sub-cover may or may not be finite.

So.

If every open cover of $X$ has a finite sub-cover, one says that $X$ is compact.

This is the closest topology allows us to get to the notion of a finite space.

That is, every subset of a compact space is of necessity compact. The converse may or may not be true. A famous example is given by Heine & Borel for $R^n$ , which I will state later.........

Ben · 06-17-2009

Hmm, losing heart here slightly. So as quick as I can....

Obviously (why?), no $R^n$ is compact, yet the Heine-Borel Thm states that the closed and bounded subsets of any $R^n$ are compact. The proof is quite hard, but uses only what we have defined so far; on current form, I am not anticipating any attempts.(Cowards!!)

Anyway. Let $X,\,Y$ be topological spaces, and let $U \in Y$ be open. Then the function $f:X \to Y$ is said to be continuous iff the pre-image set $f^{-1}(U) \in X$ is also open. (If I had thought there was genuine interest out there, I could probably have expressed this a little less compactly)

That is, continuous functions map open sets to open sets (it should be clear from our early definitions that they also map closed sets to closed sets.

And if $f:X \to Y$ is bijective (usual definition) and with a continuous inverse (same definition as above), then this function is called a homeomorphism. It is the topological equivalent of an isomorphism.

Continuous maps preserve the topological properties we(!) have been "discussing", that is, the continuous image of a connected space is connected, the continuous image of a Hausdorff space is Hausdorff, the continuous image of a compact space is compact etc.

It follows then that homeomorphic spaces are topologically indistinguishable, hence the old joke about a topologist's inability to distinguish her coffee mug from a donut.

I had planned to go from here, or hereabouts, to manifolds, then to Lie groups and their algebra. But I rather think now that I shan't......

Qfwfq · 06-18-2009

Quote:

Originally Posted by Ben

The proof is quite hard, but uses only what we have defined so far; on current form, I am not anticipating any attempts.(Cowards!!)

Actually, I remember it not being all that hard, but perhaps the fact that I can remember how it goes makes me not a valid example!

Quote:

Originally Posted by Ben

I had planned to go from here, or hereabouts, to manifolds, then to Lie groups and their algebra. But I rather think now that I shan't......

C'mon!!!

Perhaps though you are being a bit too closed and bou... ehem, compact, for some of the viewers.

Ben · 06-18-2009

Quote:

Originally Posted by Qfwfq

Actually, I remember it not being all that hard, but perhaps the fact that I can remember how it goes makes me not a valid example!

Ya, well, as has been amply documented on this site, you are a lot cleverer than I am. Suffice to say, I find it a bit tricky.

Quote:

C'mon!!!

Perhaps though you are being a bit too closed and bou... ehem, compact, for some of the viewers.

I am sure this is right, but had other viewers (were there any?) asked for expansion, I should been delighted to provide it. None did.

Don't get me wrong - I enjoy disputing with you (especially when you are wrong!), but there is little profit in preaching to the converted. Likewise, what is the point of a "tutorial" thread, whose only active viewer is one who knows the subject already?

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