I was approaching the following problem:

"Let $f \colon X \to Y$ be continuous. Is it true that if $x$ is a limit point of $A \subset X$ then $f(x)$ is a limit point of $f(A)$?"

The answer is that it is false and here is a counterexample I found: $X = \mathbb{R}$ with the standard topology, $Y = \mathbb{N}$ with the discrete topology and finally $f(x) = 1$ for every $x \in \mathbb{R}$. (it's a counter example since $2$ is a limit point of $[0,58]$ but $f(2)$ is not a limit point of $f([0,58])$. To prove this just notice that $\{1\}$ is an open neighborhood of $f(2)$ but $\{1\}\cap (f([0,58]) \setminus \{f(2)\}) = \emptyset$.

Now you are all thinking "where's the problem then?"... The problem is that my intuition failed and I tried to prove the statement for a while before trying to find a counterexample!

What can I do to avoid this problem in future? Is there any intuition I should have had to start looking for counterexamples before trying to prove the affirmative result?

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    $\begingroup$ The discrete and indiscrete topologies tend to make things break. For example, every function from the discrete topology is continuous and every function to the indiscrete topology is continuous. You can use examples based around that to try to defy how you think things should work in general. $\endgroup$ – Ian Coley Sep 15 '13 at 3:36
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    $\begingroup$ Building off Ian's comment: it is very handy in point-set topology (and everywhere else, but here in particular) to have a few go-to examples to turn to. The discrete and indiscrete topologies are such. I find that other examples like the cofinite topology, $\mathbb{R}$ with the lower limit topology, and $\mathbb{C}$ with the dictionary order topology are nice things to keep in mind here and there. As Munkres remarks, you shouldn't focus too much on weird counterexamples, but having a few go-to examples is useful. $\endgroup$ – Gyu Eun Lee Sep 15 '13 at 4:14
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    $\begingroup$ Hovever note that if $x$ is in closure of $A$ then $f(x)$ is in closure of $f(A)$. $\endgroup$ – user87690 Sep 15 '13 at 8:16
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    $\begingroup$ The constant map is always continuous and therefore is always a good choice when trying to find a counterexample. For example, assume you are supposed to verify or falsify the statement For every continuous closed map $f$ we have $f[$int$(A)]\subseteq$int$(f[A]).$ It is not immediately clear whether this should be true or not, so you could try to prove it. But once you have considered the constant map $c:\Bbb R\to\Bbb R,$ you have the solution. $\endgroup$ – Stefan Hamcke Sep 15 '13 at 16:45
  • $\begingroup$ Well, you said things that I (suppose I) am familiar with, but still... Probably I should just stop feeling ashamed and move on with my life :D thanks for all the precious recalls! $\endgroup$ – user67133 Sep 15 '13 at 19:35

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