1. ## inf(S)

) Let g : [0; 1] ! R be a continuous function. Let S be a nonempty subset of [0; 1]
and suppose that g(x) = 3 for all x 2 S. Let a = inf(S). Show that g(a) = 3

2. ## Re: need question help

I'm not sure that I understand your question, but if S is a closed set then

$\displaystyle a=\inf(S)=\min(S)$

and since $\displaystyle g(x)=3$ for all $\displaystyle x\in S$, we conclude that $\displaystyle g(a)=3$. If $\displaystyle S$ is an open set you should find a sequence of numbers that converges to $\displaystyle a$ and use the continuity property to prove that $\displaystyle \lim\limits_{x_n\rightarrow a} g(x_n)=3$ (follow this link)

3. ## Re: need question help

Originally Posted by uasac
I'm not sure that I understand your question, but if S is a closed set then $\displaystyle a=\inf(S)=\min(S)$
and since $\displaystyle g(x)=3$ for all $\displaystyle x\in S$, we conclude that $\displaystyle g(a)=3$. If $\displaystyle S$ is an open set you should find a sequence of numbers that converges to $\displaystyle a$ and use the continuity property to prove that $\displaystyle \lim\limits_{x_n\rightarrow a} g(x_n)=3$ (follow this link)
It is not the case that $\displaystyle S$ is either open or closed.
It is the case that $\displaystyle a\in S\text{ or }a\notin S$.
Now this argument works.

4. ## Re: need question help

If $\displaystyle S$ is open then $\displaystyle a\notin S$.
If $\displaystyle S$ is closed then $\displaystyle a\in S$.
I think...

5. ## Re: need question help

Originally Posted by uasac
If $\displaystyle S$ is open then $\displaystyle a\notin S$.
If $\displaystyle S$ is closed then $\displaystyle a\in S$.
I think...
What you have written there is correct.
BUT the set $\displaystyle S$ may be neither open nor closed.
It could be that $\displaystyle S=\{0.1\}\cup(0.5,1].$
That set is neither open or closed.
However, $\displaystyle \inf(S)\in S.$

It could be that $\displaystyle S=(0.1,0.5]\}\cup\{1\}$
That set is neither open or closed.
However, $\displaystyle \inf(S)\notin S.$

6. ## Re: need question help

Right... thanks for the correction.