Using Tychonoff's theorem, prove that [0,1] is compact.
Should i present [0,1] as a product of compact spaces? is that possible? if so how can I do it?
any help is appreciated!
I haven't thought about this really hard, so it might not be correct, but you can try letting under the usual Euclidean metric. Then we can represent . In other words, each copy of allows you to write down a digit of a number from . For example,
At this point, it is necessary to check that the product topology for is the same as the Euclidean topology, which I have not done. Maybe you can think about it. If it is true, then Tychonoff's theorem will finish the problem.
What is [0,1] in the context of Tychonoff's Theorem:
"If each of the subspaces is compact, the product space , with the cartesian product topology, is also compact." Taylor
Does the theorem say, for example, that (x,y,z) is a compact space if x, y, and z are? If so then (x) is compact if [0,1] is compact, which it is. But that's the opposite of what you have to prove. Stuck
I was hoping that the product topology would somehow be metrizable, maybe with a metric like (basically mimicking the Euclidean metric). It would take care of the repeated nines situation. It seems like a real hassle to deal with though!
Maybe we can simplify the situation by using the binary representation of real numbers?
Is Tychynoff's theorem a concise (abstract) way of dealing with continuity of functions of many variables (which can be quite messy)?
I would be satisfied with "yes," "no," or "irrelevant." Anything beyond that, such as its ultimate purpose (use), would be a great bonus.
Taylor states Tychonoff's theorem as:
"If each of the subspaces is compact, the product space , with the cartesian product topology, is also compact."
To me, as a special case, it says, for example, if X1, X2, and X3 are coordinatre axes (sub-spaces), then the product space (X1,X2,X3) (3d space) is compact.*
EDIT: *And then it would follow that closed subsets of the 3d space are also compact.
EDIT AGAIN: It would also make sense to me that if X1 is (x1,x2) (a plane) and X2 is (x3) A line), then X1 X X2 is (x1,x2,x3) (3d space), in the context of Tychonoff's theorem.
OK, so what does Taylor mean by "a compact subspace?" Only closed bounded spaces like [0,1]?
EDIT: I tried rereading the proof in Taylor. It's over my head. I'm gettin out while the gettins good. Sorry to bother you.