prove that [0,1] is compact?

**aharonidan** · April 5th 2011, 02:51 AM

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!
thanks

**roninpro** · April 5th 2011, 06:15 AM

I haven't thought about this really hard, so it might not be correct, but you can try letting $X=\{0,1,2,\ldots, 9\}$ under the usual Euclidean metric. Then we can represent $[0,1]\cong \Pi_{n=1}^\infty X$ . In other words, each copy of $X$ allows you to write down a digit of a number from $[0,1]$ . For example, $.14159\ldots\cong \{1\}\times \{4\}\times \{1\}\times \{5\}\times \{9\}\times \ldots$

At this point, it is necessary to check that the product topology for $\Pi_{n=1}^\infty X$ 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.

Good luck.

**FernandoRevilla** · April 5th 2011, 06:19 AM

You can identify $\mathcal{F}(\mathbb{R},[0,1])$ and the product set $F=\prod \{[0,1]_x:x\in \mathbb{R}\}$ where $[0,1]_x$ denotes a copy of $[0,1]$ .

Edited: Sorry, I didn´t see roninpro's post.

**Hartlw** · April 5th 2011, 08:21 AM

What is [0,1] in the context of Tychonoff's Theorem:

"If each of the subspaces $X_{a}$ is compact, the product space $X= \prod_{a \epsilon A} X_a$ , 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

**Opalg** · April 5th 2011, 08:54 AM

Originally Posted by roninpro

I haven't thought about this really hard, so it might not be correct, but you can try letting $X=\{0,1,2,\ldots, 9\}$ under the usual Euclidean metric. Then we can represent $[0,1]\cong \Pi_{n=1}^\infty X$ . In other words, each copy of $X$ allows you to write down a digit of a number from $[0,1]$ . For example, $.14159\ldots\cong \{1\}\times \{4\}\times \{1\}\times \{5\}\times \{9\}\times \ldots$

At this point, it is necessary to check that the product topology for $\Pi_{n=1}^\infty X$ 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.

Good luck.

That construction ought to work. But in that product topology, is it true that the sequence 0.49, 0.499, 0.4999, 0.49999, ... converges to 0.5? As usual with constructions involving infinite decimals, the ambiguity between .9 recurring and 1 causes serious headaches.

**Hartlw** · April 5th 2011, 09:15 AM

(x) is compact because R is compact. (Product space of dim 1), Tychonoffs Theorem

if (x') is a closed subset of (x), (x') is compact and hence x' [0,1] is compact.

**roninpro** · April 5th 2011, 10:37 AM

Originally Posted by Opalg

That construction ought to work. But in that product topology, is it true that the sequence 0.49, 0.499, 0.4999, 0.49999, ... converges to 0.5? As usual with constructions involving infinite decimals, the ambiguity between .9 recurring and 1 causes serious headaches.

A headache indeed.

I was hoping that the product topology would somehow be metrizable, maybe with a metric like $d(\{a_1\}\times \{a_2\}\times \{a_3\}\times \ldots, \{b_1\}\times \{b_2\}\times \{b_3\}\times \ldots)=\sum_{n=0}^\infty (a_n-b_n) 10^{-n}$ (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?

**Hartlw** · April 6th 2011, 08:41 AM

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.

**Defunkt** · April 6th 2011, 09:15 AM

What do you mean by 'dealing with continuity'? Proving that a given function is continuous? If so, then I don't really see how.

**Hartlw** · April 6th 2011, 09:40 AM

Originally Posted by Defunkt

What do you mean by 'dealing with continuity'? Proving that a given function is continuous? If so, then I don't really see how.

Defining functions in the neighborhood of a point and their limit at the point. Frankly, I looked at the discussion in Taylor and found it unintelligible, as I did previous posts, with no interest in wading through it. I would be satisfied If I could salvage a basic insight out of it.

Taylor states Tychonoff's theorem as:

"If each of the subspaces $X_a$ is compact, the product space $\prod_{a \epsilon A} X_a$ , 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.

**Defunkt** · April 6th 2011, 10:04 AM

I assume by coordinate axes you mean copies of $\mathbb{R}$ . In that case, why are you assuming $\mathbb{R}$ is compact? this is definitely not true under the usual topology.

**Hartlw** · April 6th 2011, 10:15 AM

Originally Posted by Defunkt

I assume by coordinate axes you mean copies of $\mathbb{R}$ . In that case, why are you assuming $\mathbb{R}$ is compact? this is definitely not true under the usual topology.

All singletons (x) where x is a real number. Or in the case of a plane, (x1,x2), where x1 and x2 are real numbers.

If you want to engage me in a discussion of topology, rather than answer my question, which seems straight forward, I surrender in advance.

**Hartlw** · April 6th 2011, 10:25 AM

Originally Posted by Defunkt

I assume by coordinate axes you mean copies of $\mathbb{R}$ . In that case, why are you assuming $\mathbb{R}$ is compact? this is definitely not true under the usual topology.

You are right. I admit I had trouble with the notion of points on a line being compact because of the theorem: a set is compact iff it is closed and bounded.

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.

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