# Thread: closure containing smallest closed set

1. ## closure containing smallest closed set

Prove: If $\displaystyle S \subset \bold{R}$, then $\displaystyle \bar{S}$ is the smallest closed set containing $\displaystyle S$.

So $\displaystyle \bar{S} = S \cup S'$. We want to show that if $\displaystyle T$ is any closed set and $\displaystyle S \subset T$, then $\displaystyle \bar{S} \subset T$.

So do proof by contradiction: Suppose that $\displaystyle \bar{S}$ is not the smallest closed set containing $\displaystyle S$. Is this the correct way to proceed?

2. I suggest that you start by supposing a close set T such that $\displaystyle S \subseteq T\quad \& \quad T \varsubsetneq \overline S$.
Then you know that $\displaystyle \left( {\exists x_0 \in \overline S \backslash T} \right)$. Again the complement of T is open. Get a contradiction

3. So $\displaystyle T^{c} = \overline{S} \backslash T$ is open. Suppose that $\displaystyle x_0 \in T^{c}$. Then $\displaystyle (\exists \delta > 0)[(x_{0}- \delta, x_{0} + \delta) \subseteq T^{c} = \overline{S} \backslash T]$.

But this is a contradiction because $\displaystyle T \not \subseteq \overline{S}$? So $\displaystyle T \subseteq \overline{S}$.

4. Originally Posted by particlejohn
So $\displaystyle T^{c} = \overline{S} \backslash T$ is open. Suppose that $\displaystyle x_0 \in T^{c}$. Then $\displaystyle (\exists \delta > 0)[(x_{0}- \delta, x_{0} + \delta) \subseteq T^{c} = \overline{S} \backslash T]$.
But this is a contradiction because $\displaystyle T \not \subseteq \overline{S}$? So $\displaystyle T \subseteq \overline{S}$.
No that does not work. See my suggestion.
$\displaystyle \begin{gathered} \left( {\exists x_0 \in \overline S \backslash T} \right) \hfill \\ x_0 \notin {\rm T}\quad but\quad x_0 \in \overline S \hfill \\ S \subseteq T\quad \Rightarrow \quad x_0 \notin S \hfill \\ \end{gathered}$

BTW: $\displaystyle T^{c} = \overline{S} \backslash T$ is not true.

5. shouldn't we suppose a closed set $\displaystyle T$ such that $\displaystyle S \subset T$ and $\displaystyle T \subset \overline{S}$?

Then we know that $\displaystyle \exists x_{0} \in \overline{S} \backslash T$ because $\displaystyle \overline{S}$ is "greater" than $\displaystyle T$? So $\displaystyle x_{0} \not \in T \ \text{but}\ x \in \overline{S}$. Then $\displaystyle x \in S \cup S' \implies x \in T$, which is a contradiction. Hence $\displaystyle T \supset \overline{S}$ which implies that $\displaystyle \overline{S}$ is the smallest set containing $\displaystyle S$?

6. or is this too simple?

7. That will work. But you need to suppose that T is proper subset of the closure.
That insures that $\displaystyle \left( {\exists x_0 \in \overline S \backslash T} \right)$.
Remind your reader that $\displaystyle S \subseteq T$.

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# closure of a set is the smallest closed set

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