I think he wants it proven. I don't know enough topology to know if it is obvious, but I suspect it is (as in, write down the definitions of both this example of the subspace topology and this example of weak topology, and notice they are the same).
(For instance, I don't happen to know if a 1-1 continuous map is always a homeomorphism onto its range, so I don't even know what an embedding is. :-)
@JackSchmidt Yes. So now he says this: "Since $e$ is a homeomorphism (from $X$ to $e(X)$), it is evident $X$ has the weak topology induced by $\pi_\alpha e=f_\alpha$"
@JackSchmidt The weak topology on $X$ induced by a family $\{f_\alpha:X\to X_\alpha:\alpha \in A \}$ is the topology consisting of the sets $f_\alpha^{-1}(U)$ such that $U$ is open in $X_\alpha$.
@JonasTeuwen Yes. Now recall that $$X=\prod_{\alpha\in A}X_\alpha$$ has the Tychonoff topology, and that a function $f:Y\to X$ is continuous $\iff$ $\pi_\alpha f=f_\alpha$ is continuous.
@JonasTeuwen OK. So now we have "evaluation map" $$e:X\to \prod X_\alpha$$ (here $X$ is another space, not the product from before) defined as $\pi_\alpha e =f_\alpha$, where $\{f_\alpha\}$ is a prescribed collection of functions $X\to X_\alpha$
@JonasTeuwen That if $U$ is open in $\prod X_\alpha$, then $f_\alpha^{-1}(U)$ is open in $X$. But that follows from the fact that $f_\alpha^{-1}=e^{-1}\pi_\alpha^{-1}$ both of those last are continuous, correct?
@JonasTeuwen Well, the weak topology induced by $f_\alpha:X\to X_\alpha$ on $X$ is that where open sets are of the form $f_\alpha^{-1}(O_\alpha)$ with $O_\alpha$ openin $X_\alpha$
@JackSchmidt Am I correct in saying that if $M\subset \mathbb N$ is not finite, then $\overline M=\varnothing$ and if it is finite, $\overline M =M$ (viz, the finite sets are closed).
@PeterTamaroff Close. If $M \subset \mathbb{N}$ is not finite, then $\bar M = \mathbb{N}$. The closed sets are finite or the whole set. The open sets are cofinite or the empty set.
@PeterTamaroff Weird. Yes, I think so. If both vertical and horizontal lines are open, then every point is the intersection of finitely many open sets. So every subset is a union of points, and so every subset is open.
There's a fundamental difference between the teacher understanding and the student understanding. You may have finished explaining, but there is no reason he has to be finished struggling. :-)
@OldJohn I´m having a problem to prove this(i´ts kind simple):$\overline A\capB=\overline A \cap \overline B$ intersection of the closure is the closure of intersection.
well if we have a base of cylinder set pre-images, then we showed that if x is outside of a closed set, its inside an open set, and thus inside some cylinder set pre-image, which f_i then maps to a cylinder set. we also know f_i maps B to a smaller set than the closure of f_i(B)
@PeterTamaroff and we can always find some cylinder set pre-image totally outside of B (entirely within X\B) so that f_i of this set is disjoint with f_i(B)
@PeterTamaroff figured it out? As $x \in f_{\alpha}^{-1}(V) \subset F^c$ we have $f_\alpha(x) \in V$ and $f_\alpha(F) \subset \overline{f_\alpha(F)} \subset V^c$, so $x \notin \overline{f_\alpha(F)}$.
@t.b. I'm reading Royden and when he proves that the unit ball in an infinite dimensional vector space is not compact, he seems to be implicitly using dependent choice. Do you know if DC is required?
assume x is in cl(A U B) and show if x is not in cl(A), x must be in cl(B), and then show if x is in cl(A), or x is in cl(B) that either way x is in cl(A U B)?
i've managed to count the messages by client by hour for the last several days.
but, traffic is higher in the morning than the evening and more during the week than the weekend. i'm not sure how to figure out when a drop in traffic is significiant.
@JacobSchlather I honestly don't know. I can't think of an argument that wouldn't use dependent choice for that, so all sorts of surprises are possible.
In most cases you don't need to use any choice since you have a "natural basis".
@JacobSchlather I suppose you could do something like that: take an infinite Dedekind finite set and build $l^2$ on it, and it looks like all hell breaks loose. Let me see if I find something.
@JacobSchlather Thanks, I just found something similar :) If you enter "unit ball" in the very last search field here you'll get a number of references to Brunner 1983 (a/b).
