Topology Q: Proving Local Property of Regularity, Complete Reg., Tychonoff

[cogito²]

I'm really just having trouble figuring out what a question is asking. Here's the question:

A property $$P$$ of topological spaces is said to be a local property, provided that a space $$X$$ has property $$P$$ whenever $$X$$ has a base each element of which has property $$P$$. Show that the properties of regularity, complete regularity and being Tychonoff are local properties.

My problem is really just that in proving that say regularity is a local property, I'm not sure what to use as a subspace. I could take a given base set and then consider the rest of the base sets intsersected with it as sub-base sets for that original base set. I'm not sure if that's what I'm supposed do though. Does that sound like the right approach? Any help would be swell.

 

[cogito²]

Nobody has any bright ideas :( ?

 

[NateTG]

Did you mean basis rather than base?
If you did mean 'base' then I have no clue.
If it's basis, then they seem rather straightforward:

Something like:
Assume that all basis elements are regular
Pick some $$x$$ in $$X$$.
The following holds for all basis elements $$B$$ of $$X$$:
Now, for any closed set $$C$$ that does not contain $$x$$ there is some (possibly empty) intersection $$C \cap B$$ which is closed in $$B$$ under the subspace topology. Since $$B$$ is regular, there is a set $$v_B$$ open in $$B$$ which contains $$C\cap B$$ and a set $$u_B$$ open in $$B$$ and disjoint from $$V$$ which contains $$x$$.

Then let $$U=\bigcup_{x \in B} u_B$$ and $$V=\bigcup_{x \notin B} B \cup \bigcup_{x \in B} v_B$$.
Since they are both unions of open sets, they are open. Clearly, $$C \subset V$$, and $$x \in V$$.

Of course, since this doesn't allow you to prove that the two are disjoint, you'll have to do something different. I doubt you'll have much trouble working it out.

 

[cogito²]

Yeah now that I look at it that's the way it seems I need to go. I actually got what you had after I went back and looked at it but still haven't actually been able to produce two disjoint sets. At least I now understand the problem. Hopefully I can take it from here. Thanks for the help.

 

[NateTG]

Yeah now that I look at it that's the way it seems I need to go. I actually got what you had after I went back and looked at it but still haven't actually been able to produce two disjoint sets. At least I now understand the problem. Hopefully I can take it from here. Thanks for the help.

You might consider taking closed sets in the basis sets rather than open ones...

 

[cogito²]

That homework problem is due tomorrow and I still couldn't figure it out so I went to the grader to ask him for any hints. I was not alone and when we got around to asking about that problem we found out that the grader didn't know how to prove it. That was a little disconserting but then it got a little worse when someone else showed up saying that he had just talked to the professor about it and found out that the professor didn't even know how to prove it.

Hmmmmmmmmmmmmmm. So I guess the good news is that we won't be marked down by the grader if we don't have it proven (he feels that isn't really fair since he can't prove it). It may be that the problem is not provable or it's just really, really difficult. It wouldn't be the first badly written problem in Royden (I had to prove something a few weeks back and after giving up on the proof I instead found a counter-example). If it's provable I don't think I'll be figuring it anytime soon.

So if you know a proof of this could you enlighten me? It won't give me any points but I just really want to know now. Thanks for all your help though.