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Given that $ f $ is a continuous real-valued function on $[a,b]$, then $f([a,b])$ is a closed interval. I am wondering if this fact follows for open intervals. Thanks

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    $\begingroup$ Consider constant functions. $\endgroup$ – lulu Mar 14 '16 at 13:06
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    $\begingroup$ A map which has the property that "the image of an open set is open" is called an "open map", and it is unrelated to continuity. $\endgroup$ – Patrick Stevens Mar 14 '16 at 13:09
  • $\begingroup$ Will continuous maps always yield a closed interval? $\endgroup$ – Javier Mar 14 '16 at 13:11
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    $\begingroup$ On the interval $(-1,1)$, consider $y(x)=x$ and $z(x)=x^2$. $\endgroup$ – Empy2 Mar 14 '16 at 13:12
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    $\begingroup$ The continuous image of an open interval is an interval, but the image may be open,closed, or half-open.BTW,the set $\{0\}$ is equal to the closed interval $[0,0]$. $\endgroup$ – DanielWainfleet Mar 14 '16 at 14:43
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If you have a closed (and bounded) interval $[a, b]$, then it is compact and connected. These are both properties which are respected by continuous functions in general, so $f([a, b])$ is compact and connected, and in $\Bbb R$, that means a closed interval.

For an open interval $(a, b)$, you can tell that $f((a, b))$ is connected, so it is an interval, but in general you cannot say what kind of interval (open, closed or half-open). Examples of all three:

  • Open: $f(x) = x$ for any open interval $(a, b)$.
  • Half-open: $f(x) = x^2$ for any open interval $(a, b)$ where $a<0<b$.
  • Closed: $f(x) = \sin(x)$ for any open interval $(a, b)$ with $b-a > 2\pi$.
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  • $\begingroup$ Why can't we tell what kind of interval $f((a,b))$ is? $\endgroup$ – Javier Mar 15 '16 at 4:51
  • $\begingroup$ @Javier Given a specific function, and a specific interval $(a, b)$ we can of course tell what $f((a, b))$ is. That's not what I mean. What I mean is that unlike $f([a, b])$, it can change depending on the function and interval. $f([a, b])$ is always closed, while $f((a, b))$ can be anything. $\endgroup$ – Arthur Mar 15 '16 at 6:58
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It does not, a simple counterexample is the function $f(x)=1$ with $I=(0,1)$. Here $f((0,1))=\{ 1\} $ which is not open in $\mathbb{R}$. Continuity means that the inverse image of open sets are open, hence, given any continuous function $f$ and open set $V$ in its codomain, $f^{-1}(V)$ must be open in its domain.

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With intermediate value theorem you can easily prove that if $f$ is continuous and injective, then the set $f((a,b))$ is indeed open.

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