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Sep 16, 2013 at 13:51 comment added Joseph Van Name @Niki. It will probably be best to ask an entirely new question in a new post since the new question seems significantly different from the old one.
Sep 16, 2013 at 7:28 comment added user39982 any idea? By the way is it possible to ask the question (mentioned above) in a new post? (I am new and I don't know that much about the regulations here. My new question is very similar to the new one mentioned above. so I am not sure if they accept this as a whole new question) Thanks in advance.
Sep 15, 2013 at 19:30 comment added user39982 So I have to restate my problem. I am looking for a Tychonoff space $X$ such that for any finite subset $F$, $|C(X\backslash F)| = |X| \not = |T|$ (where $T$ is the set of clopen subsets of $X$). I would appreciate your points and suggestions on this one.
Sep 15, 2013 at 18:39 comment added Joseph Van Name Yes. The idempotents in $C(X)$ are in a one-to-one correspondence with the clopen sets. Furthermore, for a compact totally disconnected space $X$, the cardinality of the collection of all clopen sets is equal to the weight $w(X)$ and is therefore bounded above by $|X|$. If $X$ is a compact ordinal space or the one-point compactification of a discrete space, then there are $|X|$ clopen subsets of $X$.
Sep 15, 2013 at 11:20 comment added user39982 Thanks to you and @JoelDavidHamkins a lot. So fruitful. Seems like a small course on $C(X)$. I must confess that I have believed wrongly that the cardinal of idempotents is equal to $2^A$ where $A$ is the cardinal of connected components of $X$ But I think the correct thing is that the cardinal is equal to the cardinal of clopen sets. Forgive me if I am asking something trivial but what is the cardinality of the set of idempotents in these four examples? Thanks anyway.
Sep 14, 2013 at 22:54 history answered Joseph Van Name CC BY-SA 3.0