Image of a close set of a continuous function is not necessarily closed Suppose a function $f: [a, b] \rightarrow \mathbb{R}$ which is continuous. What can we say on its image? I thought the image has to be closed, but it seems not true.
 A: A continuous function on a closed and bounded segment attains its minimum and maximum values and each value between them. So the image of this closed segment is again a closed segment whose bounds are the minimum and maximum of this function.
See the Bolzano's Theorem.
A: Continuous functions have a property that they map compact sets to compact sets. 
That is if $X,Y$ are metric spaces then for any continuous function $f:X\to Y$, the range of $f$ that is the set $f(X)=\{f(x):x\in X  \}$ is compact.
In $\mathbb R$, compact set is equivalent to closed and bounded set. (This follows from Heine-Borel Theorem).
So coming back to your question: $f:[a,b]\to \mathbb R$ is continuous and therefore its range that is the set $f[a,b]=\{f(x):x\in [a,b]\}$ must be compact (equivalently closed and bounded as we are in $\mathbb R$). 
So image of your $f$ can't be $\mathbb R,[0,\infty), [0,1)$ or $[1,\infty)$.
A: No. A continuous function maps compact sets to compact sets. To map to sets that are not closed you need a closed set that is not compact; in the case of the real line that means that the set has to be unbounded.
A: To give the most general answer: it does have to be closed because $[a,b]$ is a compact set, images of compact sets under continuous functions are compact (this is trivial to prove), and maps from compact to Hausdorff spaces are closed.
