) 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

Results 1 to 6 of 6

- Nov 27th 2011, 08:00 AM #1

- Joined
- Nov 2011
- Posts
- 16

- Nov 27th 2011, 08:52 AM #2

- Joined
- Nov 2011
- Posts
- 14

## 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)

- Nov 27th 2011, 09:06 AM #3

- Nov 27th 2011, 09:10 AM #4

- Joined
- Nov 2011
- Posts
- 14

- Nov 27th 2011, 09:19 AM #5
## Re: need question help

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, $\displaystyle \inf(S)\notin S.$

- Nov 27th 2011, 09:22 AM #6

- Joined
- Nov 2011
- Posts
- 14