5

Let $(X,\tau)$ be a topological space. Suppose that $B\subseteq X$ has no accumulation point. That is, for any $x\in X$ there exists some $U\subseteq X$ such that $x$ is in the interior of $U$ and $B\cap U\cap\{x\}^c$ is empty.

Claim: If $C\subseteq B$, then $C$ is closed.

Proof: If $C\subseteq B$, then, since $B$ does not have an accumulation point, $C$ does not have any, either. But any set that contains the set of its own accumulation points (which is the empty set in the case of $C$) must be closed. $\blacksquare$

Corollary If $(Y,\nu)$ is any topological space, then any function $f:B\to Y$ is continuous (in the relative topology).

Proof: If $V$ is any closed set in $Y$, then $f^{-1}(V)\subseteq B$. Therefore, $f^{-1}(V)$ is closed in the topology $\tau$ and thus must be closed in the relative topology (generated by $B$) as well. $\blacksquare$

This set of results (especially the corollary) is so surprising to me that I thought I would seek some feedback. Thank you very much for sharing your thoughts.

triple_sec
  • 23,377

1 Answers1

5

It’s entirely correct. The hypothesis on $B$ amounts to saying that $B$ is a closed, discrete subset of $X$. Every subset of a closed, discrete set is also closed and discrete, and every function whose domain is a discrete space is continuous.

A familiar example is the set $\Bbb Z$ in $\Bbb R$ with its usual topology: $\Bbb Z$ is a closed, discrete subset of $\Bbb R$, and therefore so is every subset of $\Bbb Z$. And any function $f:\Bbb Z\to Y$, where $Y$ is any space, is continuous, since every subset of $\Bbb Z$ is open.

Brian M. Scott
  • 616,228