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Yes. $I^{\alpha}=I^{\beta}$ does imply that $\alpha=\beta$. To see this, suppose that $\alpha$ is an infinite cardinal. Then each point in $I^{\alpha}$ is the intersection of $\alpha$ many open sets, …

The spaces in which every closed set is a boundary are precisely the resolvable spaces. A topological space is said to be resolvable if it can be partitioned into two dense subspaces. $\mathbf{Propos …

Since all of the answers to this question (except the one involving Alexander's subbase lemma) refer to a usually strange rehashing of the ultrafilter proof (BOO), I decided to give two nice proofs to …

I claim that for each cardinal $\lambda$, there is a connected space $C$ and $c_{0},c_{1}\in C$ such that whenever $|X|<\lambda$, then $X$ is connected if and only if for all $x,y\in X$ there is some …

Suppose $(X,d)$ is a compact metric space. Then every mapping $f:X\rightarrow X$ such that $d(x,y)=d(f(x),f(y))$ is always bijective.

We can always get non-principal ultrafilters on sets from non-discrete extremally disconnected spaces. Let $X$ be a topological space. Let $\text{Clo}(X)$ denote the Boolean algebra of clopen subsets …

It is well known that the space $\{0,1\}^{\kappa}$ satisfies the countable chain condition. Recall that a topological space $X$ satisfies the countable chain condition if and only if every collection …

We can also describe $\beta(\mathbb{Z},\mathcal{T})$ in terms of ultrafilters on Boolean algebras. I claim that $\beta(\mathbb{Z},\mathcal{T})$ is the space of ultrafilters on the Boolean algebra of c …

The kind of completely regular space you are looking for is an $F$-space. Suppose that $X$ is a completely regular space. Then we say that a subset $A\subseteq X$ is $C^*$-embedded if whenever $f:A\ri …

Let $(X,\rho)$ be a compact metric space, and let $(Y,d)$ be a separable metric space. Then I claim that one can endow $X$ with a compatible metric $d$ such that every continuous $f:X\rightarrow Y$ ca …

Since we are talking about rings of continuous functions, I will only talk about completely regular spaces in this problem. It is well known that for completely regular spaces $X$, $C(X)\simeq C(Y)$ i …

There is a nice characterization of the spaces $X$ where the Tietze extension theorem holds for all complete separable metric spaces $Y$. We say that a Hausdorff space $X$ is ultranormal if whenever $ …

It may be better for you to consider uniform spaces instead of simply topological spaces. If you have a uniform space, then there is a very natural topology that one may put on the power set. Uniform …

Following Victor Protsak's suggestion, I took the answer to this question and turned it into a paper found here Ultraparacompactness and Ultranormality, so it may be easier to read that paper than to …

I am not sure if this is the kind of answer you were looking for, but since no one has given an answer yet, I think it is a good idea to say what little I know about larger cardinal analogues of the B …