Let $(X,\tau)$ be a topological space and $A \subseteq X$. Show that the closure of $A$ is the smallest closed set containing $A$. Let $(X,\tau)$ be a topological space and $A \subseteq X$. Show that the closure of $A$ is the smallest closed set containing $A$.
Definition 1. Let $x \in X$. A set $N \subseteq X$ is called neighborhood of $x$ if there exists $U \in \tau$ such that $x \in U \subseteq N$. The set of all neighborhood of $x$ is denoted by $\mathcal{N}(x)$. Furthermore, $U$ is called open neighborhood of $x$ if $U \in \mathcal{N}(x) \cap \tau$.
Definition 2. Let $A \subseteq X$. A point $x \in X$ is called a closure point of $A$ if for any $U \in \mathcal{N}(x) \cap \tau, U \cap A \ne \emptyset$. The set of all closure points of $A$ is called a closure of $A$, denoted by $cl(A)$.
Attempt:
I want to show this only by using the definitions.
Let $A \subseteq X$. We want to show that:

*

*$A \subseteq cl(A)$;

*$cl(A)$ is closed; and

*For any closed set $D$ with $A \subseteq D$, we have $cl(A) \subseteq D$.

By Definition 2.,
$$cl(A)= \{x \in X : (\forall U \in \mathcal{N}(x) \cap \tau).U \cap A \ne \emptyset\}.$$
Is it true? If yes, is it true that
$$cl(A)^c = \{x \in X: (\exists U \in \mathcal{N}(x) \cap \tau). U \cap A = \emptyset\}$$
be its complement?
For 1: Let $x \in A$. Let $U \in \mathcal{N}(x) \cap \tau$ be given.
By definition of $\mathcal{N}(x)$, there exists $U \in \tau$ such that $x \in U \subseteq U$.
Hence, $x \in U \cap A$ and so, $U \cap A \ne \emptyset$.
Therefore, $x \in cl(A)$. Thus, $A \subseteq cl(A)$.
For 2: We'll show that $cl(A)^c$ is open, i.e., $cl(A)^c \in \tau$.
Let $x \in cl(A)^c$. Then there exists $U \in \mathcal{N}(x) \cap \tau$ such that $U \cap A = \emptyset$. By Definition 1., there exists $U \in \tau$ such that $x \in U \subseteq U \in \tau$. Hence, $cl(A)^c \in \tau$. Thus, $cl(A)$ is closed.
For 3: Let $D$ be any closed set with $A \subseteq D$. The goal is to show that $cl(A) \subseteq D$. Let $x \in cl(A)$. By definition, for all $U \in \mathcal{N}(x) \cap \tau$, we have $U \cap A \ne \emptyset$. Since $x \in U$ and $U \cap A \ne \emptyset$, then $x \in A$. Since $A \subseteq D$, then $x \in D$. Hence, $cl(A) \subseteq D$.
Therefore, by 1,2, and 3 above, we can conclude that $cl(A)$ is the smallest closed set containing $A$.
I'm in doubt whether my approach is correct. Any helps and corrections would be appreciated.
Thanks in advanced.
 A: I'm not sure about about step 2, and step 3 is definitely wrong. To be precise,

Since $x \in U$ and $U\cap A \neq \emptyset$, then $x\in A$

is wrong.

Those steps can be proved almost at the same time:
$$
\operatorname{cl}(A)^c 
= \{x \in X: (\exists U \in \mathcal{N}(x) \cap \tau). U \cap A = \emptyset\}
= \bigcup \{U \in \tau: U \cap A = \emptyset\} 
$$
Last equality is manifestation of the following: if $x\in \operatorname{cl}(A)^c $
then exists $U \in \mathcal N(x) \cap \tau$ such that $U \cap A = \emptyset$. But then for any $y \in U$, we have $U \in \mathcal N(y) \cap \tau$ and thus $y \in \operatorname{cl}(A)^c $ giving $U \subseteq \operatorname{cl}(A)^c $. Because of that, we can omit everything about neighborhoods - any non-empty open set $U$ that satisfies
$U\cap A = \emptyset$ will be added on the sum, because it is neighborhood of something. We can throw in the empty set as well, since it does not contribute anything to the sum.
Step 2  is now trivial, because we have that $\operatorname{cl}(A)^c$ is sum of some family of open sets and therefore is open.
Step 3 can be equivalently stated as $\operatorname{cl}(A)^c  \supseteq V$, where $V\in \tau$ satisfies $V \subseteq A^c$ or $V \cap A = \emptyset$.
Since complement of closure is sum of all such open sets, it naturally is the biggest one, and therefore closure is the smallest.
