Function Continuity Proof in Real Analysis

[NihalRi]

Homework Statement

We've been given a set of hints to solve the problem below and I'm stuck on one of them

Let f:[a,b]->R , prove, using the hints below, that if f is continuous and if f(a) < 0 < f(b), then there exists a c ∈ (a,b) such that f(c) = 0

Hint
let set S = {x∈[a,b]:f(x)≤0}
let c = supS, show c∈[a,b]

Homework Equations

∀ε>0, ∃δ>0 such that ∀x∈(a,b), if |x-p| <δ then |f(x)-f(p)|<ε

The Attempt at a Solution

I have not gotten very far

want to show a<c<b
c≥x ∈[a,b]: f(x)≤0
a<x<b
want to show a<x≤c<b, so show c<b because as sup it's already bigger than x
f(a) ≤f(x)≤ 0 < f(b) ⇒ f(x)<f(b)

I can see it is true intuitively but I'm struggling with showing this properly. I would appreciate your help greatly.

 

[Delta2]

It is actually easier than you thought to show that $c\in[a,b]$:
Because for every $x\in S$ we have $x\leq b$ it follows that $b$ is an upper bound of $S$, and $c$ as the supremum of $S$ is less or equal than any other upper bound of $S$ (this is a direct consequence from the definition of supremum) so it is $c\leq b$. Also easy to show that $a\leq c$

 

[NihalRi]

It is actually easier than you thought to show that $c\in[a,b]$:
Because for every $x\in S$ we have $x\leq b$ it follows that $b$ is an upper bound of $S$, and $c$ as the supremum of $S$ is less or equal than any other upper bound of $S$ (this is a direct consequence from the definition of supremum) so it is $c\leq b$. Also easy to show that $a\leq c$

Thank you so much, Can't believe it was right there and I couldn't see it. So for showing a ≤ c from the set def we know that
a≤x≤b and as c in an upper bound it's
c ≥ x and by transitivity
a ≤ a

Now on to the rest of this problem :)

 

[NihalRi]

It is actually easier than you thought to show that $c\in[a,b]$:
Because for every $x\in S$ we have $x\leq b$ it follows that $b$ is an upper bound of $S$, and $c$ as the supremum of $S$ is less or equal than any other upper bound of $S$ (this is a direct consequence from the definition of supremum) so it is $c\leq b$. Also easy to show that $a\leq c$

I can't figure out the rest of it either

The hint says show f(c) = 0 by contradiction and consider two cases. Apply the fact that if f is continuous at 'a', ∃ an open interval I centered at 'a' such that f(x) >0for all x ∈ I. Explain why 'c' not equal to 'a' which will mean a<c. Then let 'p' be any point between 'a' and 'c' which is in the I. Relate p to set S and to c = supS to show a contradiction.

So assuming f(c) not equal to 0, one case is f(c) >0
clearly as f(a) < 0, f(c) not equal to f(a) and therefore a not equal to c meaning a < c
I took 'c' to be the 'a' in the italicized part above and that the interval is (c-δ,c+δ)
then I let p be any point such that a < p < c which itself in in (c-δ, c+δ) so,
c-δ < a < p < c < c+δ and f(p) > 0, so p ∉ S and all in this interval are > 0 so not in 'I'
but this implies that the only elements that can be in 'I' are from c+δ to b which makes c no longer an upperbound but a lowerbound which is a contradiction.
I do the same for case where f(c) < 0

This does not seen right to me because I don't understand the placement of the interval. I think I misinterpreted that part but not sure what it's supposed to be. Again, I would appreciate any help.

 

[PeroK]

If we leave the formal definition of continuity to one side for now, can you see what is going on?

$f(a)$ is strictly less than 0. By continuity, $f$ cannot 'jump' immediately to be positive for $x > a$. Therefore, there must be an "interval" starting at $a$ on which $f$ is negative.

Similarly, we have an interval at $b$ on which $f$ is positive.

Now, we can turn our attention to $c$, which is strictly between $a$ and $b$.

If $f(c) < 0$, then ...?

Can you see how to proceed?

 

[NihalRi]

If we leave the formal definition of continuity to one side for now, can you see what is going on?

$f(a)$ is strictly less than 0. By continuity, $f$ cannot 'jump' immediately to be positive for $x > a$. Therefore, there must be an "interval" starting at $a$ on which $f$ is negative.

Similarly, we have an interval at $b$ on which $f$ is positive.

Now, we can turn our attention to $c$, which is strictly between $a$ and $b$.

If $f(c) < 0$, then ...?

Can you see how to proceed?

I see the logic of the first statements, here's my attempt
There is also a possibility that there are several c's for which f(c) = 0 correct? But we are trying to show if there is at least one.

If $f(c) < 0$, then it's in an interval (a,c]
I am thinking this is where p comes in and there is an open interval around any p with f(p)<0
sorry if I'm still approaching this all wrong

 

[PeroK]

I see the logic of the first statements, here's my attempt
There is also a possibility that there are several c's for which f(c) = 0 correct? But we are trying to show if there is at least one.

If $f(c) < 0$, then it's in an interval (a,c]
I am thinking this is where p comes in and there is an open interval around any p with f(p)<0
sorry if I'm still approaching this all wrong

What's $p$?

$c = sup \{x \in [a, b]: f(x) \le 0 \}$

So, there is only one $c$.

There could be any number of points where $f(x) = 0$. But, that's not the issue. The issue is to show that $f(c) = 0$.

 

[NihalRi]

What's $p$?

$c = sup \{x \in [a, b]: f(x) \le 0 \}$

So, there is only one $c$.

There could be any number of points where $f(x) = 0$. But, that's not the issue. The issue is to show that $f(c) = 0$.

$c = sup \{x \in [a, b]: f(x) \le 0 \}$
f(c) < 0
c ∈ [a, b] a≤c≤b
a≤x≤b
c ≥ x
x≤c≤b

f(b)>0
b∉S

Intuitively it makes so much sense and I feel really close but it hasn't clicked yet​

 

[PeroK]

$c = sup \{x \in [a, b]: f(x) \le 0 \}$
f(c) < 0
c ∈ [a, b] a≤c≤b
a≤x≤b
c ≥ x
x≤c≤b

f(b)>0
b∉S

Intuitively it makes so much sense and I feel really close but it hasn't clicked yet​

I can't make any sense of those steps. You could try the following problem:

Let $f$ be continuous on $(a, b)$. Let $c \in (a, b)$ with $f(c) > 0$.

Show that there exists an open interval $I \subset (a, b)$, with $c \in I$ and:

$x \in I \ \Rightarrow \ f(x) > 0$

That result is a fundamental building block of the proof required for your problem here and many others. If you are stuck on the bigger problem, try this one, which is an important component of the proof you need.