Let $X,Y\subset \mathbb R$. Is it true that $f:X\to Y$ continuous $\iff$ $f^{-1}(U)$ open for all open subset $U\subset Y$. The implication is of course correct. But I have doubt for the reciprocal. I know that for all $U$ in the topology of $Y$ the result is true (it's in fact the topological definition of continuity). But I suspect that there it can have set $V$ open for the topology of $Y$ that is not open in $\mathbb R$ such that $f^{-1}(V)$ is not open in $X$, and thus that $f$ wouldn't be continuous. What do you think ?

  • 1
    $\begingroup$ If this is not your definition of continuity, then what is it? $\endgroup$ – Magdiragdag Sep 16 '17 at 9:33

You right ! Let $$f:[0,1]\to [0,1]$$ defined by $$f(x)=\begin{cases}0&x\in [0,1)\\1&x=1.\end{cases}$$ You will have that $f^{-1}(U)$ is open for all open $U\subset [0,1]$ but $f$ is not continuous.

  • 2
    $\begingroup$ Note that $(\frac12,1]$ is open in $[0,1]$, but its preimage $\{0,1\}$ is not open in $[0,1]$, so $f$ is not continuous. $\endgroup$ – Magdiragdag Sep 16 '17 at 9:34
  • 1
    $\begingroup$ @Magdiragdag: Yes, but since the tag is written as "real-analysis", I suspect that the OP is asking about open of $\mathbb R$ that are included in $[0,1]$. $\endgroup$ – Surb Sep 16 '17 at 9:35
  • $\begingroup$ That is probably the OP's source of confusion but the OP does state "[there is a] set $V$ open for the topology of $Y$ that is not open in ${\mathbb R}$", so the OP is not thinking about that. $\endgroup$ – Magdiragdag Sep 16 '17 at 9:37
  • $\begingroup$ @Magdiragdag: Oh yes, you right... but Surb : it's exactly the type of example I was looking for (I indeed confused open of $\mathbb R$ included in $Y$ and element of $Y$ topology... I have so much problem with this things....grrrrr :-) Topology is so hard to give natural representation...) $\endgroup$ – MathBeginner Sep 16 '17 at 9:41
  • $\begingroup$ @MathBeginner: At the beginning it looks strange, I know. But don't worry, you will catch it :) $\endgroup$ – Surb Sep 16 '17 at 9:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.