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Here's a proof that if $X$ is any simply connected Hausdorff space such that $X\setminus \{p\}$ is path connected for all $p\in X$, then the complement of any totally disconnected subset is connected. …

No. There are locally connected subsets of $\mathbb{R}^2$ which are totally path disconnected. See my answer to this old MO question "Can you explicitly write R2 as a disjoint union of two totally pat …

There is a simple proof that a game of Hex must have a winner, which implies the result you want.See here: Brouwer's Fixed Point Theorem and the Jordan Curve Theorem, Lemma 5.5. The Brouwer fixed poin …

No, $S$ does not have to span $C(X)$. Taking the case with $X=[0,1]$, let $\mu$ be any atomless finite signed measure whose positive and negative parts $\mu^+$,$\mu^-$ have full support, so that $\mu …

No it is not possible. Suppose that $X\times Y\cong\mathbb{R}^n$. Then, as the product is contractible, both $X$ and $Y$ must be contractible spaces. For any $x\in X$, I'll show that $\lbrace x\rbrace …

To answer the main question -- there are no nontrivial self-isometries of $\mathcal{GH}$. I can give a proof of this, but as it is getting rather long, I will state some facts in $\mathcal{GH}$ witho …

Yes, it is possible to prove that $f^{-1}(0)$ has an unbounded component. I'll give a proof of this. I see that BS already has one proof -- maybe that can also be translated into a similar argument. I …

No, it is not possible for $\mu$ to be complete. There exists a closed subset $K$ of $X$ with $\mu(K)=0$ and a continuous onto map $f\colon K\to2^\omega$. With $K,f$ as above, if $A\subseteq …

Let $S$ be a subset of the reals such that $S \cap [a,b]$ and $S^c \cap [a,b]$ cannot be written as a countable union of closed sets for any $a < b$. This can be done (this explicit example of a non-B …

There's already been some good answers to this. However, this is something that I have also thought about recently, because I happen to have come across several meagre sets of full Lebesgue measure in …