inf(S)

**sbacon1991** · November 27th 2011, 08:00 AM

) 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

**uasac** · November 27th 2011, 08:52 AM

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

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

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

**Plato** · November 27th 2011, 09:06 AM

Originally Posted by uasac

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

It is not the case that $S$ is either open or closed.
It is the case that $a\in S\text{ or }a\notin S$ .
Now this argument works.

**uasac** · November 27th 2011, 09:10 AM

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

**Plato** · November 27th 2011, 09:19 AM

Originally Posted by uasac

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

What you have written there is correct.
BUT the set $S$ may be neither open nor closed.
It could be that $S=\{0.1\}\cup(0.5,1].$
That set is neither open or closed.
However, $\inf(S)\in S.$

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

**uasac** · November 27th 2011, 09:22 AM

Right... thanks for the correction.

Thread: inf(S)

LinkBack

Thread Tools

Display

inf(S)

Re: need question help

Re: need question help

Re: need question help

Re: need question help

Re: need question help

Search Tags