no, that's not right, @Studentmath.

How come?

Given $x$, there is $\epsilon>0$ so that ... what?

So now I've figured that @HipsterMath is an old friend, eh?

mmmmm chocolate moose

no, chocolate steer

Wait, $A$ is nowhere-dense iff $IntClA=\emptyset$ by definition, right?

But you messed up your negations ...

I don't know if I'm allowed to say, but she was once Charlie, @Ted

Yes, of course, I know, @Mike.

Ah... alright.

Wanna come practice deboning a pork loin and butterflying it, @Mike?

You just want me for slave labor!

Well, that's what you're worth.

Harsh... but fair.

I'm not sure I follow - $ClA=A$, so $IntA=\emptyset$, $x\in IntA$ iff there is $\epsilon>0$ so that $B(x,\epsilon)\subseteq A$. So it's sufficient to show that for every $\epsilon, x\in A$, $B(x\epsilon)$ will have an element of $A^c$? Or am I messing it up?

Pork!

I can't remember all these sign conventions @Ted

Oh dear, @Studentmath. Nowhere dense should mean that for every $x\in X-A$, there is some $\epsilon>0$ so that $B(x,\epsilon)\cap A =\emptyset$.

You're at a Republican convention again, watching signs, @Mike?

I hate being 'lost in translation'. Perhaps the definition I am looking for is meager, then?

I think meager = nowhere dense. @Mike?

Yes, @Ted, and the Laplace operator is the worst candidate of all!

Well, what's a minus amongst friends?

Meager = countable Union of nowhere dense

Wish I didnt remember that.

This readily enters the area of topology that makes me hold my nose.

Oh, bah, first and second category ... I don't think that's germane, @Studentmath.

Hush, @Mike ... You're fast approaching Balarka territory.

No doubt about that.

sigh I will just stick to the given definitions, call it Jacob as far as I care :P

Jacob's a nice name for a subset of a topological space.

I think I prefer Paul for this situation, though.

Surely it should be easy to see what I said works, @studentmath.

Take an $x\in\ell^2$, $x\notin \prod [0,1/N]^N$. Find a neighborhood disjoint therefrom.

Just so long as you rob Paul to pay Peter @Mike

Ah, well, that's simple. There is some $x_n\in x$ so that $x_n >1/n$. I take $\epsilon=x_n - 1/n$, and it will be disjoint

Yes, I just learned about that inequality @Ted

Good, @Mike ... I think I only saw it referred to as such in Warner, although I'd forgotten.

Sounds good, @Studentmath.

Cheers :)

OK, movie time.

OK, good excuse for me to depart, too.

And I'm off to bed, g'night and thanks!

3 people just left, lol.

I was looking at some questions given in some math contest for high school. Here is another one ... (very nice)

Things died down rather quickly in here...

I am having dinner again

Too hungry

@teadawg1337 I am still around to talk.

It's already Christmas Eve here.

Do you celebrate Christmas, @Jasper?

This is an exercise from 'Special Functions' be George Andrews. This is the first step in a proof of Euler's reflection formula, but I'm struggling on where to begin.

I'm given the hint let $g(x) = \frac{d^2}{dx^2} \log \phi (x)$

@KhallilBenyattou Nope. But I will celebrate the day I start studying math again, which is hopefully next Thu.

and observe that $g(x+1)=g(x)$ and $\frac{1}{4}(g(\frac{x}{2})+g(\frac{x+1}{2})) = g(x)$

@KhallilBenyattou Yeah, I've tried that. It's probably the route I'm supposed to take, I just haven't managed to get anywhere with it.