Proving a set is closed

**p00ndawg** · November 16th 2009, 11:42 AM

Let S = [0,4]. Show S is closed.

How can I show that the bd S is a subset of S, or that R/S is open?
I know that simply stating the bd should be enough to say that S is closed, but my professor wants us to be able to also show bd{0 , 4}.
I know that if it were (0,4) I could say that (0 - E, 0+ E) intersected with S and R/S is non empty, but is it sufficient to say that if S is closed then the bd S = bd (R/S) so R/S is empty?

Also at what point when x is in bd does (x - E, x + E) intersected with S and R/S non empty? This would only be true for an open set correct?

**Plato** · November 16th 2009, 11:49 AM

Originally Posted by p00ndawg

Let S = [0,4]. Show S is closed.

This is the complement: $\left( { - \infty ,0} \right) \cup \left( {4,\infty } \right)$ .
Can you prove that the union is open?

**p00ndawg** · November 16th 2009, 12:54 PM

Originally Posted by Plato

This is the complement: $\left( { - \infty ,0} \right) \cup \left( {4,\infty } \right)$ .
Can you prove that the union is open?

yes i think i can, especially since I know that the union of any collection of open sets is open.

so by showing that the complement is open, we can say that the set is closed.

nice, thank you.

**srinivas.sista89** · November 16th 2009, 11:40 PM

by bd(s) do you mean the boundary of S ??

then bd(S) = S closure intersection (X-S) closure. and since S is closed S=S closure so you are done.

**redsoxfan325** · November 17th 2009, 12:20 AM

Originally Posted by srinivas.sista89

by bd(s) do you mean the boundary of S ??

then bd(S) = S closure intersection (X-S) closure. and since S is closed S=S closure so you are done.

The point was to prove that $S$ is closed.

**Drexel28** · November 17th 2009, 06:33 AM

An alternative way to do this (which is just implicitly noting that $\mathbb{R}-S$ is open) is to suppose that $\ell\notin[0,4]\implies \ell\in\mathbb{R}-S$ . We have two cases, either $\ell<0$ in which case note that $\ell+\frac{0-\ell}{2}<0\quad\color{red}\star$ ; or $\ell>4$ in which case $\ell-\frac{\ell-4}{2}>4\quad\color{red}\star\star$ . And clearly $\ell-\frac{0-\ell}{2}<0\quad\color{red}\star$ and $\ell+\frac{\ell-4}{2}>4\quad\color{red}\star\star$ . If the first case is two then choosing $\varepsilon=\color{red}\star$ shows that $\mathcal{U}_{\varepsilon}\subset (-\infty,0)$ and if the second is true then choosing $\varepsilon=\color{red}\star\star$ shows that $\mathcal{U}_{\varepsilon}\subset(4,\infty)$ . In either case $\ell$ is an interior point of $\mathbb{R}-S$ . Thus $\mathbb{R}-S$ is open which implies that $S$ is closed.

Thread: Proving a set is closed

LinkBack

Thread Tools

Display

Proving a set is closed

Similar Math Help Forum Discussions

proving a set is closed

Proving a set is closed and convex.

Proving a set closed

Proving the 'Closed Ball' is closed

Proving the Zero Set of a Continuous Function is Closed

Search Tags