2:28 PM
A problem on locally compact spaces in main chat room:
in Mathematics, 39 mins ago, by user193319
Problem: Let $X$ be some locally compact space, $Y$ some generic topological space, and $f : X \to Y$ a continuous open map. Show that $f(X)$ is locally compact. Here is my proof: Let $y \in f(X)$. Then there exists an $x \in X$ such that $f(x) = y$. Since $X$ is locally compact, there exists $C \in X$ compact such that $x \in U \subseteq C$, where $U$ is some open nbhd of $x$. Then $y = f(x) \in f(U) \subseteq f(C)$. Since $f$ is open, $f(U)$ will be open;
in Mathematics, 39 mins ago, by user193319
...and since $f$ is continuous, $C$ will be compact. This shows that $f(X)$ is locally compact.
in Mathematics, 14 mins ago, by user193319
@Daminark Do you know a little topology? If so, would you mind checking the proof I gave above?
in Mathematics, 11 mins ago, by Daminark
Sounds good to me

2:41 PM
in Mathematics, 8 secs ago, by Martin Sleziak
@user193319 I will just remind this about locally compact spaces: "There are other common definitions: They are all equivalent if X is a Hausdorff space (or preregular). But they are not equivalent in general." BTW feel free to post also in general topology chat room - perhaps you could help to bring new life there; it was rather inactive for some time.
BTW some posts on the main:

in Mathematics, 1 min ago, by Secret
Ok, I now know that all ordinals < $\omega_1$ are totally disconnected, but what about ordinals after $\omega_1$ (under the order topology), are they also totally disconnected?

@Secret So you are asking whether order topology on any ordinal $\alpha$ is totally disconnected?

yeah, and in fact, naively, I am wondering whether all well ordered sets are totally disconnected since the fact that we can always find the minimum element for every subset, should mean they are all singletons and thus any two elements can never be partitioned in terms of union of disjoint open sets?
My attempt so far is to choose $(0,\beta)\cup (\beta,\infty),\beta \in \alpha$, analogous to how total disconnectedness is proved for the rationals, but then I always miss 0 and $\beta$

So you are saying basically this: If we have any set $A$ and $m=\min A$, then both $\{m\}$ and $A\setminus\{m\}$ are clopen in $A$?
Hence if $A$ has at least two elements, then it is not connected.
@Secret Why don't you take $[0,\beta+1)$?
For order topology, the sets of the form $[0,x)=(-\infty,x)$ belong to the standard base, too.

2:56 PM
ah, I thought too much about "what should I do with all the negative numbers"
I suspect that is what I try to express when you elaborated in term of the set $A$
right, if I use $[0,\beta+1)\cup (\beta,\infty)$ then I have successfully partition $\alpha$ into two disjoint open subsets thus (I think I need to double check to ensure I did not mix up totally disconnected with connected)

Not just that. You can partition any subset $A\subseteq\alpha$ in such way. (If $A$ has at least two elements.)

I see, thus since I can do the same partition throughout all ordinals and subsets of them, it follows that all well ordered sets are totally disconnected if it has 2 or more elements

Yes, I believe it's correct.

hmm, now it makes me wonder, whether the class of all ordinals is also totally disconnected since it is well ordered, but I need to check whether various notions of topology can generalise to proper classes
4

Topological spaces are usually modelled over sets. What happens if I try to topologize a proper class? Say, the class of all maps from naturals to cardinals or similar. Suppose I specify a topology on such a class by defining convergence for a net. What usual properties or facts may I lose? I wou...

Here is a question on the main, but it did not get much response: Topology in a proper class.
Almost at the same time :-)

3:09 PM
yeah, had to be very careful when playing with these really huge objects

7 hours later…
9:58 PM
Hello. I am working on the following problem: Show that $[0,1]^\omega$ is not locally compact in the uniform topology. I have been able to reduce the problem to showing that $\overline{B}(0,\epsilon)$, the closure of the $\epsilon$-ball, is not compact. The uniform topology is induced by the metric $\rho(x,y) := \sup \{ \min \{1, |x_i-y_i| \} ~|~ i \in \Bbb{N} \}$.
Since we are dealing with a metric space, I compactness of closed $\epsilon$-ball is equivalent to limit point compactness and sequential compactness. I having trouble deciding which route to take to obtain a contradiction. In the one case, I need to find an infinite set in $C = \overline{B}(0, \epsilon)$ that has no limit point, or I need to find a sequence in $C$ that has no convergent subsequence. However, both seem equally difficult. I could use a hint.