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I have recently stumbled across a proof in a book about functional analysis concerning closure of the set and I can't seem to grasp one part of it.

Let $X$ be a normed space over $\mathbb{K}$ and $M$ be a subset of $X$. Then, we define the closure $\overline{M}$ of $M$ as $$ \overline{M}:= \text{smallest closed set of $X$ containing $M$}. $$ Then, the proposition states that for $M$ being a nonempty subset of $X$, the following holds:

$(i)$ $u\in \overline{M}$ iff, for some sequence $\{ {u_n}\}$ in $M$ $$ u_n \to u \quad \text{as} \quad n\to \infty. $$ In the proof the author denotes $C$ as the set of all points $u\in X$ such that $u_n\to u$ as $n \to \infty$ for some sequence $\{ {u_n} \}$ in $M$. Then, he shows that $C$ is closed and that part of the proof is clear to me. However, for the converse implication he proceeds as follows:

Let $\mathcal{C}$ be a closed subset of $X$ such that $M\subseteq \mathcal{C} $. Then $$ C \subseteq \mathcal{C}. $$ Therefore, $C$ is the smallest closed subset that contains the set $M$, i.e., $C=\overline{M}$.

I can't see where does the statement $C\subseteq \mathcal{C}$ come from and would be very thankful for any explanation of it.

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  • $\begingroup$ What is your starting point? What is the def'n of "closed set"? $\endgroup$ – DanielWainfleet Apr 2 '17 at 12:00
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Let $(X, \|\cdot\|)$ be the normed space. You want to show that $C \subseteq \mathcal C$.

Let $u \in C$. Then you know there exists a sequence $\{u_n\}$ in $M$ such that $u_n \to u$.

Suppose for the sake of contradiction that $u \not \in \mathcal C$, that is $u \in X \setminus \mathcal C$. But $X \setminus \mathcal C$ is open (since it's the complement of a closed set) and hence there exists $\varepsilon > 0$ such that $B_\varepsilon \subsetneq X \setminus \mathcal C$. For this $\varepsilon > 0$ there exists $N_\varepsilon \in \mathbb{N}$ such that for all $n \geq N_\varepsilon$ you have $\|x_n - x\| < \varepsilon$, that is $x_n \in B_\varepsilon(x)$. But since $M \subseteq \mathcal C$ you have $M \cap B_\varepsilon(x) = \emptyset$ and this contradicts the assumption $x_n \in M$ for all $n \in \mathbb{N}$.

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Take $x \in X\backslash\mathcal{C}$. There exists an open ball $B \ni x$ such that $B\cap M = \emptyset$. Therefore $x\in X\backslash C$ and we have shown that $C \subset \mathcal{C}$.

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