Show that the closure of a set A is the smallest closed set containing A.

I need to prove that $\bar A$ (the closure of set A) is the smallest closed set containing A. We have already proved that it is a closed set, so now I just need to show that it is the smallest one.

I have written a preliminary proof that I don't think is particularly rigorous, and would be grateful if someone could give me some pointers. We haven't covered anything regarding metric spaces or anything else in topology, so I had trouble understanding other solutions posted on the site.

Let $A$ be a non-empty set, and $\bar A$ the closure of $A$ (the union of $A$ and all of its limit points). Let $B$ be a closed set with $A \subset B \subset \bar A$ and $B \neq \bar A$.

Since $A \subset B$ and $B \subset \bar A$, $B$ consists of all the elements of $A$ and some (but not all) of its limit points. However, this means that there are sequences contained entirely within $B$ whose limit points are not elements of $B$. Thus, $B$ is not closed, posing a contradiction to the original statement.

Solution 1:

I think it is clearer without a contradiction argument:

Let $B$ be a closed set satisfying $A\subset B$. Let $a$ be a limit point of $A$, so that we can find a sequence $(a_{n})_{n}\subset A$ with $a_{n}\rightarrow a$. Since this sequence is also contained in $B$ and $B$ is closed, it follows that $a \in B$. Because $a$ was an arbitrary limit point, $B$ contains the closure of $A$.

Solution 2:

Say $\bar A = \cap \{ F \supseteq A : F \text{ closed}\}$. We know $A$ is a closed. If $x \in \bar A$ then there is sequence $(x_n)$ with ${x_n} \to x$. Then ${x_n} \in F$ but also $F$ closed set. So $x = \mathop {\lim }\limits_{n \to \infty } {x_n} \in F$. That is $\bar A \subseteq F$. So $\bar A \subseteq \cap \{ F \supseteq A:F \text{ closed}\}$

Solution 3:

Yes, your proof is quite correct. Maybe one suggestion would be to make it clear immediately that you are going for a contradiction.

... Let Suppose that there is some closed set be $B$ be a closed set with such that $A \subset B \subset \bar A$ and $B \neq \bar A$. ...