@PeterTamaroff probably more useful to concentrate on compactness - a much more widely useful concept

@JasperLoy yep

@JasperLoy I will get to connectedness in a while.

CH4

Compactness is CH5

Yes, sometimes sequences don't capture the topology of a space, and you must cast a wider net to find an appropriate object that does.

@JasperLoy Bleh =)

@HenryT.Horton nice pun :)

The logical structure of physics.

@JasperLoy good - but not quite as subtle :)

@JonasTeuwen A unicorn galloping through a rainbow

@PeterTamaroff Yeah! Bring it on!

@JasperLoy Eventually physicists will realize there is no golden urn at the end of the rainbow. And neither there are strings.

(Whatever those are)

My health is good, my weapon is sharp, and my armor is shining brightly. Bring it on!

Or maybe not.

@JasperLoy Ask Jonas. I have 3D glasses!

overdose of Jura here :)

@JasperLoy Points are zero-dimensional, he could already see them before

@JasperLoy Dude, WTF?

@HenryT.Horton HAHAHAHHAHAHAHAHAHAH

@JasperLoy Gut für dich.

@Gigili I just thought of the funny picture of a woman shaped armour. Like that one in Alice in Wonderland.

Takes the awesomeness of the thing, IMO.

Said the philosopher.

@JasperLoy I can imagine a universe in which all the laws of physics are different from the ones we know ... but primes are still the same :)

HELP HELP!

@PeterTamaroff shoot

"Again, sorry, but this Mendelson book is just too STUPID!" whoever starred this:

motores.com.py/foro/index.php?attachments/…

@OldJohn Its about the product topology

Let $\{(X_\alpha,\mathfrak I_\alpha)\}$ be an indexed family of topological spaces.

Why do you use $\mathfrak{I}$ for a topology man...

I like $\tau$

How do you guys define the product topology?

@anon I'll give you some $\tau$.

In terms of the union of intersections of preimage of open sets under the projection maps?

mmm, yummy $\tau$

The coarsest topology such that the projections onto each factor are continuous

Like, how do you index them?

@HenryT.Horton Sure sure.

But how do you index them?

Like $$\bigcap_{\alpha \in A} \prod_{x\in X} p_x^{-1}(O_\alpha)$$?

yes, the open sets should be unions of intersections of preimages of open sets under the projections

The book is using some notation I can't get.

Not that one, of course.

But something like

I'm not sure how to interpret what you've written

@anon Yes, I know, sorry.

The book writes something like this:

WTF!? The projections are also labeled by $\alpha \in A$... $p_\alpha : \prod_{\beta \in A} X_\beta \longrightarrow X_\alpha$

an open set should be of the form $$\bigcup_{i\in I}\bigcap_{\alpha \in A}p_\alpha^{-1}(O_{\alpha,i})$$

An open set...

nevermind.

@anon OK, yes. The book writes.

consisting of the unions of all sets of the form $$p_{{\alpha _1}}^{ - 1}\left( {{O_{{\alpha _1}}}} \right) \cap \cdots \cap p_{{\alpha _k}}^{ - 1}\left( {{O_{{\alpha _k}}}} \right)$$ with $${O_{{\alpha _k}}} \in {\mathfrak I_{{\alpha _k}}}$$ and $i=1,\dots,k$

Why finite $k$?

Do you know what a subbase for a topology is, Piotr?

oh, you're right, it should be finite

@HenryT.Horton Not really. I know what a base is, though.

You want finite intersections because infinite intersections of open sets might not be open

@anon I don't really know what the hell I wrote there, in retrospective.

that means that the base for the product topology is given by products of sets from the component spaces, such that all but finitely many of the factors are the whole topological space and the rest are open sets

if it helps to think that way

@anon OH!!!!!!!!!!!!!

I see.

Because

In the finitary case, we have this

$$\prod\limits_{i = 1}^n {{O_i}} = \bigcap\limits_{i = 1}^n {p_i^{ - 1}\left( {{O_i}} \right)} $$

Convince yourself that in order for the projection maps to be continuous, we need these to be open sets in the product topology, and that no more open sets are required (hence the "coarsest" description)

@anon Yes, I know why that is the case.

That is because we have

goddamn 8 red coins, where are you

$$p_i^{ - 1}\left( {{O_i}} \right) = {X_1} \times \cdots \times {X_{i - 1}} \times {O_i} \times {X_{i + 1}} \times \cdots \times {X_n}$$

@anon You want to buy a can of beans?

Peter: yep

(at least in the finitay case)

Now I want to order myself with the arbitrary case

Why is it the case the index set is finite?

I don't understand the given indexation, really.

Convince yourself that if $f_i:A\to B$ is continuous for each $i\in I$ and $\cal B$ is a basis of open sets for $B$, then $f_i^{-1}(O)$ is open in $A$ for each $i\in I,O\in \cal B$, and hence the topology generated by finite intersections and arbitrary unions (see: subbase) of the inverse images of open sets of $B$ forms a subtopology of $A$. If we want the coarsest topology on $A$ for which the $f_i$ are continuous, then this is the topology.

The idea you need to grasp it what it means for a set of subsets of $X$ to generate a topology, or equivalently for a topology to be the coarsest one containing a given set of subsets of $X$. This topology must contain the subsets and be closed under arbitrary unions and finite intersections, by definition, so we simply take the set of all subsets of $X$ generated that way from the given subsets and we end up with a topology.

In the finitary case, where the family is indexed by $I=\{1,\dots,n\}$ the topology consists of sets of the form $$\bigcup\limits_{\alpha \in A} {\prod\limits_{i = 1}^n {{O_1}} } = \bigcup\limits_{\alpha \in A} {\bigcap\limits_{i = 1}^n {p_i^{ - 1}\left( {{O_i}} \right)} } $$

Where $A$ is some indexing set.

@anon What is a "subtopology"?

I understand what is going on, I just don't understand the notation...

(I guess) =P

If $\tau$ is a topology on $X$ and $\rho\subseteq\tau$ is a subset that is also a topology on $X$, I call it a subtopology.

I wonder where and when things happen that I don't notice. Everybody has everybody's skype.

@anon OK.

@Peter, You were #3 in the election or four?

@Gigili Really? I don't-

@Gigili Nou clue! I got pens an shirts though!

Umm.

@anon I kinda get that.

@Gigili I'm chatting with everyone but you on Skype right now.

@HenryT.Horton Really? I should get a skype account then!

I don't get the notation of this $$p_{{\alpha _1}}^{ - 1}\left( {{O_{{\alpha _1}}}} \right) \cap \cdots \cap p_{{\alpha _k}}^{ - 1}\left( {{O_{{\alpha _k}}}} \right)$$

OH! Wait!

means $\alpha_1,\cdots,\alpha_k$ are $k$ elements of $A$

@anon Yes! I got it now!

@anon So those sets are just the product of some open sets, and the rest are the whole space (which is also open)

We could use a countable amount but this one is the weakest of all the topologies

So we're indeed good with this one.

yes

@anon Good, good.

Thanks, dude.

I'll be off now.

Could I get your Skype usernames, peoples of the math?

I don't use Skype.

OK. Nevermind.

I thought the Skype thing was serious.,