MHB Understanding the Proof of a Bounded Function in a Closed Interval

[Petrus]

Hello MHB,
I have hard understanding a proof..
"show that if a function f is cotinuetet in $$[a,b]$$ Then f is limited."

pretty much I Dont get the poin, I got the proof in swedish but Dont understand what is happening.. Any advice or Link that explain this proof well?Regards,
$$|\pi\rangle$$

 

[Plato2]

Re: Help understand a proof

"show that if a function f is cotinuetet in $$[a,b]$$ Then f is limited."

The correct statement is, if a function $$f$$ is continuous on a closed interval then it is bounded there.

The proof really depends upon what you know.
Do you know that a closed interval is compact?

Do you know that if a function $$f$$ is continuous on a closed interval then it is uniformly continuous there?

Tell us what you have to work with.

 

[alyafey22]

Re: Help understand a proof

Do you know that a closed interval is compact?

I think you mean closed bounded interval because $$[a,\infty)$$ is closed but not bounded hence not compact.

EDIT : I know a proof using the Bolzano-Weierstrass theorem if you took it.

 

[Plato2]

Re: Help understand a proof

I think you mean closed bounded interval because $$[a,\infty)$$ is closed but not bounded hence not compact.

That depends upon how one uses the term interval. As I use it $$[a,\infty)$$ is not an interval.
I understand an interval as a bounded connect set.

 

[Petrus]

Re: Help understand a proof

Thanks evryone for taking your time! Did take me a lot of time but I got it now!:)
Regards,
$$|\pi\rangle$$

 

[alyafey22]

Re: Help understand a proof

Thanks evryone for taking your time! Did take me a lot of time but I got it now!:)
Regards,
$$|\pi\rangle$$

Can you sketch a proof ?

 

[Petrus]

Re: Help understand a proof

Can you sketch a proof ?

Ehmm.. My proof is wrong.. I am back to square 0.. I Will post later the proof That i want to try understand (it's on swedish I Will need to translate)
Regards,
$$|\pi\rangle$$

 

[Plato2]

Re: Help understand a proof

Ehmm.. My proof is wrong.. I am back to square 0..

Look into your notes/text to see if you have done these theorems. If not try them.

1) Every sequence contains a monotone subsequence.
2) Every bounded monotone sequence has a limit point.

Now the interval $$[a,b]$$ is closed. The limit of any convergent sequence from the set is in the set.

Suppose that the function $$f$$ is not bounded above on $$[a,b]$$.

That means the exists a sequence of points from $$[a,b]$$ such that $$\forall N~[f(x_N)>N]$$.

Because of continuity if $$(y_n)\to L$$ then $$f(y_n)\to f(L)$$.

There is a lot left out of that. But does that help you?