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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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    thanks alot Brian!Can you please explain me about the completely regular property?I meant the general idea.How can I find a function which maps to 0 and 1??2012-11-05
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    @ccc: In this case you can’t: if there were one, $f^{-1}\left[\left[0,\frac12\right)\right]$ and $f^{-1}\left[\left(\frac12,1\right]\right]$ would be disjoint open sets in $X$, one containing $a$ and the other $x$, and such sets don’t exist in this space. I’m not sure how much you already know; for a start you might read [this](http://en.wikipedia.org/wiki/Tychonoff_space).2012-11-05
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    @ Brian, can you please explain me how to prove that X is regular and completely regular,If X is the discrete topology,and Indiscrete topology..2012-11-07
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    @ccc: Discrete: For any $x\in X$ the function that sends $x$ to $0$ and every other point of $X$ to $1$ is continuous, so $X$ is comp. reg., and $\{x\}$ and $X\setminus\{x\}$ are disjoint open sets separating $x$ from any set not containing it, so $X$ is reg. (Remember, in the discrete top. **every** set is open.) Indiscrete: If $x\in X$, there is no non-empty closed set not containing $X$, so $X$ is vacuously reg. and comp. reg. (The only closed sets are $\varnothing$ and $X$, since those are the only open sets.)2012-11-07
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    Thank you very much Brian!I really appreciate your response..:)2012-11-07
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    @ccc: My pleasure; glad it helped.2012-11-07
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    @ Brian,this is with reference to your statement "Assume now that A is a proper subset of X, let x∈X∖A, and let....."here,since U∈τ,we have {x} is closed since U=X∖{x}.So cl{x}={x}.Am I correct?But what will be the set V∈τ such that a∈V⊆clV⊆U???Can you please explain me???2012-11-07
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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