Rudy the Reindeer

Me? Yes, I have several accounts.

I'm just thinking along.

Sorry, didn't mean to interrupt.

What's $i$?

Then let's do it the other way. I'm happy with either way.

Yes I will. I just tried it again and still can't get it right. But I'm sure I can find the mistake later.

@0celo7 No problem, glad I could help!

@0celo7 Okay, I think the answer is no. $V$ is open so $V^c$ is a closed set containing $U$. Hence $\overline{U}$ is contained in $V^c$.

Sorry again ^.^;

Ah not disjoint.

@0celo7 You implicitly require $U \cup V$ to be the whole space? That's impossible in a connected space. But if you don't require that you could to something like $U=[-1,0)$ and $V=({1\over 2}, 1]$ in $[-1,1]$ with the standard topology.

(I don't have a quibble with that, just need to be clear) Oh, okay.

Do you use $\subset$ to mean $\subsetneq$?

@0celo7 What prevents you from taking $A=B$?

Oh, hello there mean square.

@0celo7 I'm starting to think that maybe the answer is no.

Sorry. We have to try again.

It's not.

oops, I didn't check if this space is connected.... let's see.

$$ T = \{ \varnothing, \{a\}, \{b\}, \{a,b\}, \{b,c\}, \{a,b,c\} =X\}$$ with $U = \{a\}$ and $V=\{b,c\}$. Then the closure of $U$ is $\{b,c\}$ so they are not disjoint.

Ok, so how about this:

Don't worry I'm just thinking aloud.

No, not possible to find disjoint open subsets.

How about the finite complement topology on $\mathbb R$?

Good question, let's see...

It's to anyone here in the room. Not you in particular, don't worry about it.

Am I missing something? It seems to me that $A\cap\left\{ \left(p-r,p+r\right)-\left\{ p \right\} \right\} \neq\emptyset$ is exactly what needs to be shown. And nothing more.

Funnily, I posted it about 4 minutes ago and he's already accepted it.

Today I was so desperate to find something I like answering that I answered my first multivariable calculus question.

@J.M. Hehe, good reputation you have on the mathematica site.

I'm going afk for a while. See you later! Good to see you J. M.!

Especially if, once you hit the post button, a previously posted full answer appears that is better than yours : D

That is reasonable : D

@J.M. I wish it did for me. Sometimes I want to answer something and I go there and nothing happens. : D As if someone froze the front page.

Sometimes I just browse the front page for stuff I might want to answer.

@J.M. Me too. I also miss better quality of questions and better quality of answers.

Not much going on here, is there?

@J.M. Haha, that is the pot calling the kettle black! How are you? : )

This = Possible duplicate of Determining the number of subgroups of $\mathbb Z_{14} \oplus \mathbb Z_6$? I think the answer by Henning Makholm is really good. Much better than all the others.

Well I guess unless you figured out the answer in the meantime.

@JessyCat It was a nice question. No need to delete it.

Spend long time writing up nice explanation that explains everything but the very last step. Then godzilla tramples into china shop and just posts the last step.

shrugs

Nah. And it's probably coincidence.

Interesting. I leave a comment on one of BD's answers and hours later I get a downvote on one of my top answers (40+ upvotes)... Coincidence?

In real life you don't collect stupid mistakes like that either.

I think the delete button should permanently delete rather than hide answers.

I just wrote a shitty answer. I hate writing shitty answers. : (

See you all later!

@DanielFischer Well, no, not an ad. It's... decorative, I guess. But thank you for being disturbed by it too : D