The definitions I'm working with:

$(X,\mathcal{T})$ is said to be completely regular if for every $x \in X$ and every closed set $C \subseteq X$ not containing x can be separated by a continuous function.

$x$ and $A$ can be separated by continuous function if $\exists$ a continuous function $f : X \rightarrow [0, 1]$ such that $f(x) = 0$ and $f(a) = 1 \ \forall a \in A$.

I'm having trouble showing that every completely regular space is regular. Let $(X,\mathcal{T})$ be a completely regular space, and let $x \in X$ and $C$ be a closed set s.t $x \not\in C$. Then we have $f(x) = 0$ and $f(C) = 1$ and $f(x) \cap f(C) = \emptyset$ since $x \notin C$.(Not sure about the part that follows) Pick disjoint open sets U, V around $f(x)$ and $f(C)$ respectively, then $f^{-1}(U) , f^{-1}(V)$ are disjoint open sets around x and C, and hence, we have that $(X,T)$ is regular.


What you’ve done is fine apart from a small notational glitch: $f[C]$ is by definition the set of values $f(y)$ for $y\in C$, so it’s $\{1\}$, not $1$. And if you want to talk about $f(x)\cap f(C)$, you need to view $f(x)$ as a set as well, which means that you’re really looking at $f[\{x\}]$, the image of the set $\{x\}$.


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