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This might be a silly question, but if I have a function defined at only one point. Is the function continuous at that point?

Rrmm
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1 Answers1

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Depends on the topology of your spaces. Take an open set $V$ in your target space containing the image of the point and consider $f^{-1}(V)$. This will consist of the single preimage point where f is defined. Then $f$ is continuous (at the point where it is defined), if the singleton point is open in the topology of the initial space $X$, for $f: X \rightarrow Y$.

Gary.
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  • If $V \subset Y$ is open in $Y$, then $f^{-1}(V)$ is either $\emptyset$ or $X$, by definition they are open. (Actually $X$ has only one topology) So there is no need to discuss topology. – MonkeyKing Apr 28 '15 at 02:48
  • But $X$ is not necessarily a 1-point space; we may have, e.g., the Real line, but $f$ is defined at just one point. – Gary. Apr 28 '15 at 04:03
  • Well, then that is not a function. See the third example in the right column. – MonkeyKing Apr 28 '15 at 04:08
  • Well, many things can happen. I thought the OP may have been thinking of functions at isolated points in some subspace. I am trying to find a reasonably interpretation to an ambiguous/incomplete question. – Gary. Apr 28 '15 at 04:11
  • His/her wording is clear enough, that this function is defined at only one point. It is a well-posed question. In sense of math if you guess OP may expressed differently as s/he thought, it is better to put this guess as side-remark. – MonkeyKing Apr 28 '15 at 04:20
  • OK, whatever, I am just trying to find a way of treating the question so that it is not completely trivial, so the thread can have some value other than $f^{-1}(x)$ is open by default. Maybe generalizing, or creating broader context, which many have expressed they appreciate. – Gary. Apr 28 '15 at 04:25