@DavidZ That kind of thing is a good target for the "destroy user" feature. I'm told it feeds into a network wide ban algorithm of some kind, and that user had many accounts.

I wish there was, even for <2k.

You are taking about this right?

Really marvel at the user from the title itself in the URL bar, without having even the slightest idea what the body of the Q was.

But it is amusing to imagine what that would be :P

Yeah, seriously!

Am I wrong here?

reddit.com/r/explainlikeimfive/comments/2hlpld/…

@ChrisWhite Oh screw you! :P Screenshot PLIZ? It'll take me about a year or at least half of it, at this rate.

Can anyone help me with this 'visualization' which has me thoroughly confused? It appears in Reed & Simon's book on Functional Analysis, in the chapter on Topological spaces. Link to follow shortly

this statement about the cylinders is just ??? for me

@Danu I believe these are cylinders in a very loose sense - there's no restriction of the components in these directions, and in finitely many dimensions, the vectors are confined to a circle - and the ordinary cylinder is $\mathbb{R} \times S^1$, i.e. infinite in one direction, but confined to a circle in the other

Ok, so let me get this straight

@Danu You really shouldn't strive to reach 10k for that post - I could have very well lived my life without ever reading it.

We have infinitely many ($n=1,2,\dots$) neighborhoods

each one adding another $\psi_i$ (to what??)

The first one is $N(\psi_1,\epsilon_1)=\{\phi\biggl|\bigl| (\psi_1,\phi)\bigr| <\epsilon_1\}$, right?

@ChrisWhite I am not sure whaty ou're saying

I am really at a loss with this construction, still

and they use it repeatedly, so I really need it ;)

The $\psi_i$ are not orthogonal, right

Well, the $\lvert(\psi_i,\phi)\rvert < \epsilon_i$ says the component in the $\psi_i$ direction may be at most $\epsilon_i$ (which, for small $\epsilon_i$, one could call "orthogonal enough", I guess)

also, what the hell is $\phi$ (or $\varphi$)

All $\phi$ fulfilling the relation are members of the neighbourhood

waaait

okayy

the $\phi$ are what the neighborhood is actually composed of

of course

this is starting to make some sense

But now I'm not sure anymore why they would call that a cylinder - if you do it in 2D, and choose $n = 1$, you get a strip of width $2\epsilon$ around the origin.

so, in finitely many dimensions (the span of $\psi_1,\dots,\psi_n$), $\phi$ have to be 'small'

If you do it in 3D, and choose $n = 2$, then you get a rectangular prism

That's not really a cylinder...

but you have to choose all n

$n$ is running from $1,2,\dots$

Yes

is my earlier statment correct, btw?

I just tried to get a real cylinder in a special case to see why one might generalize to call this kind of neighbourhood a cylinder

@Danu Which one?

the one about the span

Yup, I think so

urgh

this is just so completely unintuitive

@ChrisWhite Yeah, topologically, they're not distinct

@ACuriousMind isn't the relationship stronger than that?

Some kind of limit...

yeah

But, in spaces with norms, I'm used to distinguish circles and rectangles, so I'm not really sure why they'd like to call that a cylinder

Ahhhh

What I don't understand about sentence about the existence of $M^\perp$ (which is basically the subspace spanned by the $\psi_1,\dots,\psi_n$). How can we even discuss the subspace spanned by the $\psi_1,\dots,\psi_n$ when we're talking about a running index $n$ here instead of a fixed number $n$??

That meant as "for every fixed $n$, the following statement holds", I think

So, for every neighbourhood $N$, there is such a space $M_N$.

Okay

Would the following be a more intuitive statement than the textbook:

Thus, the neighborhoods in the weak topology are 'restricted to be small' in finitely many dimensions (namely, the number of dimensions of the subspace spanned by the $\psi_1,\dots,\psi_n$)

In the other dimensions, they are unrestricted

I'd agree with that, but it can well be that we are missing something ;)

DAMNIT

haha

I understand that you can't commit to this haha

seriously though

calling something a 'cylinder in a dimension' if it is unrestricted in that dimension... really?

I've understood now :D At least, the last sentence makes a lot of sense now (because $N$ is unrestricted in the directions that $M$ extends into)

NOW, does anyone have an intuitive understanding how this is a neighborhood base?

@Danu Observe that the neighbourhood $N(\psi_1,\dots,\psi_n;\epsilon_1,\dots,\epsilon_n)$ is mapped to the open set of numbers smaller than $\sum_i \epsilon_i$.

sure

So it makes the scalar product continuous

Hm

Wait, that's not actually true

Why should any neighborhood of $0$ contain one of these $N$

I think I give up

^classic math

@Danu All too true :D

sad truth. I think I've come to the conclusion that I *(&*ing hate topology already :)

This is the ugly kind of topology

I like these nice and sweeping results like the Riesz-Markov theorem (positive linear functionals $\Leftrightarrow$ measures)

but I don't think it's enough :P

empty mod flag queue, wheeee :-)