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Let $f:[0,1]to mathbb R$ be continuous and such that $f(0)<0$ and $f(1)=0$. Show that the set $x in (0,1] : f(x)=0 $ has a minimum.

So I though because the function is continuous, it is bounded. Since it is bounded it must have a inf and a sup. Since a inf exists in a bounded set I could achrimedian proof it is the minimum. But my professor told me that since the set he wants a minimum in is unbounded my logic wont work. So what am I doing wrong?

The point of the exercise is this:

The set $x in (0,1] : f(x)=0 $ obviously has an infimum, because it is non-empty and bounded below. But is this infimum necessarily a minimum? That is, does the infimum itself belong to the set? This is what you have to prove.

It seems to me that you missed this. It also seems that you must have misunderstood your professor -- what he allegedly said makes no sense to me..

(Also, I presume you mean "archimedean proof". But what do you mean by this?)

Continuity and $f(0)<0$ implies that there exists a small $epsilon>0$ such that if $0leq x<epsilon$, $f(x) < 0$. Then, the zeros of $f$ are contained in the compact set $[epsilon,1]$. Now let $x_0 = inf_[epsilon,1]left x : f(x) = 0 right$. If this is a finite set, then we are done. If not, consider a sequence from the set converging to $x_0$. Since $f$ is continuous, $f(x_0)=0$ and certainly $x_0in[epsilon,1]$ due to compactness. Hence $x_0$ is the desired minimum.

Hint $:$ $x in (0,1] : f(x) = 0 = f^-1 (0).$ Since $f$ is continuous so $f^-1 (0)$ is a closed subspace of $[0,1]$ (Since $0 $ is a closed subset of $Bbb R$). Since $[0,1]$ is a closed subspace of $Bbb R$ with usual topology so $f^-1 (0)$ is a closed subspace of $Bbb R$ with usual topology.