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" The empty set and all subsets of $X$ containing a fixed subset $A$ , form a topology on $X$ "
What do you think about this statement?

Is this regular or completely regular?

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    What do *you* think about the statement? What have you tried so far?2012-11-05

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$\newcommand{\cl}{\operatorname{cl}}$We have a set $X$ and a subset $A\subseteq X$, and we define $\tau=\{\varnothing\}\cup\{U\subseteq X:A\subseteq U\}$; this is certainly a topology on $X$. If $A=\varnothing$, $\tau=\wp(X)$ is the discrete topology, and $X$ is even metrizable, so it’s certainly regular and completely regular. If $A\ne\varnothing$, however, matters are very different.

First, if $A=X$, then $\tau$ is the indiscrete topology. This is regular, though not $T_3$ (= regular plus $T_1$) or completely regular unless $|X|=1$; I’ll leave the easy proofs to you.

Assume now that $A$ is a proper subset of $X$, let $x\in X\setminus A$, and let $U=X\setminus\{x\}$; clearly $U\in\tau$. Let $a\in A$; then $U$ is an open nbhd of $a$. Can there be a $V\in\tau$ such that $a\in V\subseteq\cl V\subseteq U$? Note that in that case $\cl\{a\}\subseteq U$; now what is $\cl\{x\}$ in this topology?

Finally, note that $\{x\}$ is a closed set not containing $a$; can there be a continuous $f:X\to[0,1]$ such that $f(a)=0$ and $f(x)=1$? What would $f^{-1}\left[\left(\frac12,1\right]\right]$ be?

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    @ccc: Just take $V=\{a\}$; remember, **every** subset of $X$ is both open and closed, so $a\in\{a\}=\operatorname{cl}\{a\}\subseteq U$.2012-11-07