# Proving a set of a subset of the range of a function has a maximum

I have a homework question which is:

$f(x)$ is continuous on $(0,1)$ , $\underset{x\to 0+}{\mathop{\lim }}\,f(x)=-1$ and $\underset{x\to 1-}{\mathop{\lim }}\,f(x)=1$

Let $A=\{x\in(0,1)|f(x)=0\}$. We will mark $\operatorname{sup} A = s$.

prove that $f(s)=0$.

Obviously this means we need to prove that $A$ has a maximum. I find this fairly reasonable and intuitive and I think I might have written a correct proof but I am sure there is a much better way to do it then what I have tried.

EDIT: This is what I did roughly lined out:

1)Because of the limits exists a left area of 1 where $f(x)>0$ and a right area of $0$ where $f(x)<0$

2) choose $y_0$ and $y_1$ in those areas so and use the interval $[y_0,y_1]$ to show that $A$ is a subset of $[y_0,y_1]$ which is a subset of $(0,1)$

3)$y_1$ is that maximum of $[y_0,y_1]$ but does not belong to $A$ because $f(y_1)>0$ so choose the first $x$ in $A$ which is smaller then that and that is the maximum of $A$.

4) because $A$ has a maximum then s is equal to it and then also $f(s)=0$

What do you guys think?

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Show us your proof then. – Graphth Dec 9 '11 at 15:32
@Graphth I'd prefer not to it's very long and I would have to translate it. I am certain there is a better way to do this so I'd rather not get of track – Jason Dec 9 '11 at 15:42
Note that you could define $g(x)$ on $[0,1]$ with $g(0)=-1$, $g(1)=1$, and $g(x)=f(x)$ on $(0,1)$. Then the limit conditions of $f$ are true if and only if $g$ is a continuous function on $[0,1]$. It's not clear why the problem isn't stated in terms of a continuous function $g$ on $[0,1]$ rather than as a problem about a continuous function $f$ on $(0,1)$. – Thomas Andrews Dec 9 '11 at 15:50
@ThomasAndrews yep I could do that - I can also prove that A is a subset of some other closed interval inside (0,1) but how would that help me? – Jason Dec 9 '11 at 15:53
It helps because if $A$ is closed in a closed subset of $\mathbb{R}$ then it is also closed in $\mathbb{R},$ while the same is not true for open subsets. $(0,1)$ is closed in $(0,1)$ but certainly not in $\mathbb{R}.$ – Ragib Zaman Dec 9 '11 at 15:57

Note that the assumptions about your function imply that $A\ne\emptyset$. (Do you know why?)

There are two possibilities:

• Either $s\in A$ and then clearly $f(s)=0$.

• Or $s=\operatorname{sup} A\notin A$.

Can you show in the second case that there is a sequence $(a_n)$ of elements of $A$ such that $a_n\to s$? Can you use this finish the proof? (Does the fact that $f$ is continuous help? If $f$ is continuous and $a_n\to s$, what can you say about the sequence $f(a_n)$?)

Other possibility would be to show that the set $A$ is closed. As $A\subseteq[0,1]$, it is also bounded. Hence $A$ is compact and every compact subset of $\mathbb R$ has a maximum.

To be sure that you understand the proof, it might be useful to you to try to find an example of function, which is not continuous and for which the claim fails. (It may help you understand better at which point of your proof continuity was used.)

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As for the first point you made - Yes I know why (It was a part of the question which I had no trouble with - so I omitted it). As for your guidance on the question no sequence comes to mind and I don't understand what that sequence should help me with :S – Jason Dec 9 '11 at 15:41
To show the existence of such sequence, try to use the definition of supremum. When you have such sequence, you could use continuity. (Of course, it depends on how much you've already learned about continuous functions.) – Martin Sleziak Dec 9 '11 at 15:45
BTW have you learned already some basic facts about compact (=bounded and closed) sets? That would be other way to go. – Martin Sleziak Dec 9 '11 at 15:46
I have learned a little bit of facts about compact sets and tried using them for a while before I gave up on that approach. As for the sequence I don't understand how that sequence is supposed to help me so I don't know what sequence I am supposed to prove exists – Jason Dec 9 '11 at 15:49
You want to show that if $\sup A$ is not in $A$ then there is at least some sequence in $A$ whose limit is $\sup A.$ Basically this says that even though the supremum is not in $A$, elements in $A$ get very very close to this number (arbitrarily close in fact). Of course, this must be true if you think about what the definition of supremum is. The reason this helps is because continuous functions only change a small amount if you move a small amount. So if $f$ is zero at all these points arbitrarily close to the supremum, it must be zero itself. – Ragib Zaman Dec 9 '11 at 15:55

By the limit conditions, there exists $\epsilon > 0$ such that $A \subseteq [\epsilon, 1-\epsilon].$ Consider $g$, the restriction of $f$ onto this set. Since $g$ is continuous and $\{ 0 \}$ is a closed set, $g^{-1} ( \{ 0\} ) = A$ is closed in $[\epsilon, 1-\epsilon],$ i.e $A$ contains the limit of any sequence in $A$, and since $\sup A$ must be the limit of some sequence in $A$ we have $\sup A \in A.$

This would be the other way to go, as Martin said in the comments to his answer.

A useful lemma for the elementary supremum properties is this: Let $A$ be bounded from above so that $s=\sup A$ exists. If $t$ is a real number such that $t< s,$ then there exists an element $a \in A$ such that $t<a.$ Using this fact we can inductively construct a sequence of elements in $A$ which tend to $\sup A$. Let $a_1$ be an element such that $s-1 < a_1.$ Let $a_2$ be an element $a_2 > \max \{ s-1/2, a_1 \}$, and $a_3$ be an element such that $a_3 > \max \{ s-1/3, a_2 \}$ and so on. Can you see why i) we can always pick such elements, ii) this sequence is increasing and bounded, and thus has a limit, iii) why the limit is $\sup A$ ?

Once you know there is a sequence of elements in $A$ which tends to $\sup A$, use this characterization of continuity (if you haven't this before it is a good exercise to try to prove it): If $f$ is continuous and $a_n \to L$ then $\lim_{n\to\infty} f(a_n) = f(L).$

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