# Thread: closure containing smallest closed set

1. ## closure containing smallest closed set

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

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

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

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

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

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

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

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

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

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

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 $\left( {\exists x_0 \in \overline S \backslash T} \right)$.
Remind your reader that $S \subseteq T$.

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### prove that the closure is the smallest closed set

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