Recent content by jeckt

yeah thanks dick. I was just trying to get the ideas down but mathematically I'd have to define it much more rigorously. Other than that I think that's all the compact subsets there are for this question. Thanks for the great help guys.

oh wow! Thanks HallsofIvy! That really clears things up - so any set that contains it's limit points and finite subsets of \mathbb{Q}\cap [0,1] are compact. So the set you mention doesn't have its limit point; which is zero. If the set contained 0 then any cover of the point 0 will contain...

Been so busy lately haven't been able to give this some thought. Hmmm I get how the topology works now (i think) but the problem becomes i can't think of a subset of \mathbb{Q} \cap [0,1] that isn't compact. Maybe you can head me in the right direction. And again - thanks for all the help

Sorry I'm a bit slow but I'm not seeing how open intervals in \mathbb{R} \cap [0,1] \cap \mathbb{Q} has an infinite number of points (you mean infinite number of rational points right? if that what you mean then that makes sense to me)

Hmmm...I'm not seeing why the set would be compact. Maybe I'm viewing the relative topology wrong. I'm thinking that the relative topology is rational points or sets of rational points in [0,1] as the above posters have said. So the set you said - there wouldn't be a finite subcover the way...

I'm also doing this question right now - and the answer aren't very clear but I agree with the OP. The only compact sets are finite subsets of \mathbb{Q}\cap [0,1] since the relative topology is rational points. But that seems too simple...

Yeah I totally get you micromass! been so busy - haven't been able to reply. So gathering all that together the compact subsets are thus: \emptyset For a_{k},b_{k}\in \mathbb{R} \mbox{ s.t. } a_{k}<b_{k}\ \forall k \mbox{ and } n\in \mathbb{Z}^{+} \mbox{ s.t } n<\infty...

Thanks micromass! As for your original questions...I'll have a think about it - but from a quick glance, it seems that the -\infty plays a large roll in being able to have compact sets of the form (a,b].

What is ]a,b]? notation, I've never seen it before. As for the second point, wouldn't a subcover such as (-\infty , a) where a is any number greater than 0. Sorry about this -I'm really slow when it comes to maths.

Homework Statement Identify the compact subsets of \mathbb{R} with topology \tau:= \{ \emptyset , \mathbb{R}\} \cup \{ (-\infty , \alpha) | \alpha \in \mathbb{R}\} . just need help on how would you actually go about finding it. I usually just find it by thinking about it. The Attempt at a...

Thanks dextercioby, yeah I checked some online resources and textbooks. Just that in my analysis class the lecturer used a strict subset or I wrote it down wrong but it's all good now and I get where the issue is. Thanks for the help guys!

This makes more sense now, thanks guys! the reply is so quick too. In class, the definition I had was that a subcover is a strict subset of a cover. So in this case the subcover would be the cover and thus it would be finite. @gb7nash - yeah I know that to show a set is compact i need to show...

Hi All, This is really a stupid question...I can't seem to get my head around it and it's making me depressed just thinking about it. Anyways, let's consider \mathbb{R}^{n} the set S = [0,1] is not compact (I know it is but I can't see the flaw in my argument which seems it should be...

Fixed it up: Let A be the set of all finite sets, by the axiom of pairing there exists sets \{ A \}, \{ \{ A \}, A \}. Apply the axiom of regularity to \{ \{ A \}, A \}. There are two elements in \{ \{ A \}, A \}, consider the element \{ A \}. Clearly \{ \{ A \}, A \} \cap \{ A \} \neq...

...that makes life so much easier! (well that's if my solution works now) I've been stuck on this question for over a week - thinking about it every now and then. So gather from what you hinted. 1. Let A be the set of all finite sets 2. Then \{A\} is a finite set and \{A\}\in A 3. Apply the...