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If $f:[0,1]\rightarrow [a,b]$ is a continuous function, and $f([0,1])=(c,d)\subset [a,b]$. Is $f^{-1}([a,b])$ open or closed in $[0,1]$?

If open: Since $[a,b]$ is closed so is $f^{-1}([a,b])$; contradiction.

If closed: Since $(c,d)$ is open so is $f^{-1}([a,b])=f^{-1}((c,d))$; contradiction.


Def. for a continuous function in general topology:

Let $X$ and $Y$ be topological spaces. A function $f : X \rightarrow Y$ is continuous if $f^{-1} (Y)$ is open/closed in $X$ for every open/closed set $V$ in $Y$.

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    $\begingroup$ Why do you think you have a contradiction? $\endgroup$ – Michael Albanese Jun 20 '15 at 2:48
  • $\begingroup$ $f^{-1}([a,b])=[0,1]$ so it's both open and closed. $\endgroup$ – David Peterson Jun 20 '15 at 2:48
  • $\begingroup$ @Michael Albanese: Because when I suppose it to be closed/open it comes to be open/closed! $\endgroup$ – L.G. Jun 20 '15 at 2:49
  • $\begingroup$ @HIP13044b: A set can be both open and closed, which is the case here. In fact, for any topological space $(X, \tau)$, $X$ is always open and closed (as is $\emptyset$). $\endgroup$ – Michael Albanese Jun 20 '15 at 2:51
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The image of a closed bounded interval under a continuous function is closed and bounded (follows from Bolzano intermediate value theorem, and the fact that f atteins a min and a max, in $\mathbb{R}$)! Further, in the image of any compact (ie sequentially compact) metric space is also compact. How can you have a continuous function with this property? If a function satisfies $f([0,1])=(c,d)$ then it's not continuous.

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  • $\begingroup$ Thank you very much! Finally I got it. $\endgroup$ – L.G. Jun 20 '15 at 3:12
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If $f:[0,1]\rightarrow [a,b]$ is a continuous function, and $f([0,1])=(c,d)\subset [a,b]$. Is $f^{-1}([a,b])$ open or closed in $[0,1]$?

The answer is that it is closed and not open. It is also open and not closed. It is also neither open nor closed. It is also both open and closed. How can this be?

Well, the problem is that you have assumed that $f$ is a function that cannot exist. In particular, if $f$ is a continuous function on $[0,1],$ then since $[0,1]$ is compact, we have that $f\bigl([0,1]\bigr)$ is also compact, so will be open only if it is empty.$ However, the image of a non-empty set cannot be empty!

Why can we draw all of the strange conclusions mentioned above, then? The Principle of Explosion.

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