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I was reading the answer to this question:

Can locally "a.e. constant" function on a connected subset $U$ of $\mathbb{R}^n$ be constant a.e. in $U$?

and understand everything except the last line: "By the connectedness of $U$, it follows that $W_{43}=U$"

I don't understand how we use the connectedness here and what is the set $W_{43}$. Can someone explain this to me?

peastick
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1 Answers1

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$W_{43}$ is just a misprint for $W_m$. I believe it is also technically incorrect that $W_m=U$.

However, we can understand the proof as follows. Since $W_m=W_k$ or $W_m\cap W_k$ for each pair of indices $m$ and $k$, we can pass to a subsequence so that all the $W_m$'s are pairwise disjoint. Now $\{W_m\}_{m\in I}$ (where $I$ is an index set) is a disjoint open cover of $U$. Since $U$ is connected, the only such disjoint open cover consists of a single set. Now $U\subset W_m$ which means $U$ is constant a.e.

Ben W
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