By the way, the answer given in the paper linked to in the dual modules question is quite nice, but I really don't understand how passing to the dual category makes this argument any more transparent.
I've done that, yes, and I'm pretty sure I received quite a few of such votes, too. But in this case I really fear that eng will suffer a heart attack.
Will this suffice? To prove that the topology $\mathcal{T}$ generated by the basis $\mathcal{B}$ is equal to the union of all elements in $\mathcal{B}$?
Now the second question I have: Given a collection $\mathcal{B}$ with the following property: If $B_1, B_2 \in \mathcal{B}$ then for every $p \in B_1 \cap B_2$ there exists $B_p$ such that $p \in B_p \subset B_1 \cap B_2$ then $\mathcal{B}$ is the basis of some topology $\mathcal{T}$. Can you describe $\mathcal{T}$?
@ZeeshanMahmud If you have the bookmark installed, you need to click it each time you enter chat. Otherwise you need to install it
@BenjaminLim You're not wrong, but it suffices to take $\mathcal{T}$ to be all possible unions of elements of $\mathcal{B}$. No need to take intersections.
so in $\mathbb{R}$ it suffices to show that for every open interval $(a,b)$ and a point $x \in (a,b)$ there is an interval $(c,d)$ where $c,d \in \mathbb{Q}$ such that $x \in (c,d) \subset (a,b)$
@BenjaminLim I guess I should point out one subtlety here. The lemma you stated required you to check for every open set $U$ in $\mathbb R$. But you started off with "it suffices to check for every open interval $(a,b)$ and..." Do you see why it suffices?
Ya. In fact, the open intervals form a basis for the euclidean topology, that's why. This is one advantage in keeping the basis as small as possible. You need to check smaller number of things.
In other words: for every $p \in U$ there's some inerval $(a,b)$ such that $p \in (a,b) \subset U$ and now you can take $(c,d)$ with rational endpoints with $p \in (c,d) \subset (a,b)$.
@BenjaminLim oh, that was just a silly allusion to my agents down under.
So, a set in $S \in \mathcal{S}$ is either of the form $U \times Y$ for some open $U \subset X$ or of the form $X \times V$ for some open $V \subset Y$
@tb There are two meanings, which -- I think -- is causing some confusion to Ben: (a) It is the intersection of all topologies that contain $\mathcal S$. (b) it is the union of finite intersections of members of $\mathcal S$.
Ok, for now how about this perhaps simpler problem, if $\mathcal{B}$ is a basis for a topology $\mathcal{T}$ on X, then the topology generated by $\mathcal{B}$ is equal to the intersection of all topologies on $X$ that contain $\mathcal{B}$
Many are inexplicably defensive about it. I feel like it has eaten at Shimura for most of his life, for example, and he's a professor at Princeton. How can you still feel small?
@robjohn I think if I could choose freely I'd do part 3 in Cambridge and then try to get a PhD position there. I really like England and Cambridge is lovely.
@robjohn I worked there for a while for a games company as a Junior SWE. The housing is not so nice but I found English people, at least the ones at work, nice.
Jech is a good book as a reference. Personally, of course, I much prefer the Boolean-valued models approach to forcing, but in practice you don't really work with that. So the approach in Kunen might be useful for modern forcing techniques.
In Kunen, $p \leq q$ reads as $p$ is stronger than $q$ and in the lecture it reads as $p$ is weaker than $q$. I wonder why they do it one way or the other. It seems to make no difference so why not stick to the book.
It's the Jerusalem tradition vs. the rest of the world.
I may be in troubles if I'll do my Ph.D. with Magidor. He's in Jerusalem, so the Jerusalem notation is appropriate, but he himself switched to the worldly notation... so I should follow the place or the advisor??
I also wonder why all the letters have to be either curly, have a tilde or have a dot somewhere. I don't think it looks pretty. Why not just use regular letters like e.g. $U$ instead of $\mathcal{U}$ to denote an ultrafilter?
@Matt I don't care, as long as the introduction section says "We shall denote $p\le q$ as $p$ being stronger than $q$", or replace "stronger" by "extends".
@JM Of course. I have too much backbone. I even stand up to people that I don't have to stand up to, or that I know from the beginning will rule me over and make me change things.
This is one of the reason I do my best to drink daily. Alcohol is known as "courage juice"!
Also @JM: If you can put the MathJax link (as starred by Bobby J.) in a titled form (like I did with the ChatJax link) I'll pin it to the top.
Speaking of drinking, I am out of beer $\stackrel{\circ\cdot\circ}{\frown}$
