# If every subset of $M$ is clopen, then show that any function $f: M \rightarrow N$ is continuous where M and N are metric spaces

Here is the reasoning I think fleshes things out more. Some say that the fact follows obviously from the definitions (which may be true), but students often want an answer that explains in some level of detail.

What do you think of the following answer? Does this explain enough or are there areas that need more elaboration. Any comments are appreciated.

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Let $f: M \rightarrow N$ where M and N are metric spaces (or Topological spaces)

By the topological defintion of continuity, $f$ is continuous iff the pre-image of each open(closed) set in N is also open(closed) in M.

Now, pick any open subset $X \in N$, the pre-image of $N$ under $f$ must be open since we are given that every subset of $M$ is open (since it is clopen). QED.

One can run the same argument with closed sets.

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Isn't that the argument one has in mind when one says "folows obviously from the definition"? And you may want to replace metric with more general topological spaces. –  Hagen von Eitzen Oct 25 '12 at 18:23
@HagenvonEitzen: yes, but I have had plenty of my friends come to me and ask why and how. For some, it may be difficult to see how it follows from definition chasing; for others, it is natural. Regardless, mathematics should be made clear whenever possible. I just thought it would be nice to just flesh things out. I generally don't like to "keep in the head" and say "obviously" if I have a chance to just say it. if you have the argument in mind, then either say it or explain briefly how that happens. Thanks, I will make the proposed edit. –  user43901 Oct 25 '12 at 18:27