Different proof that $\overline{A\cup B}\subset \overline{A}\cup \overline{B}$

Question

As Paul Erdos once said, first find a proof that works, and then go looking for the beautiful proof.

Before looking at the proof my textbook gives for $\overline{A\cup B}\subset \overline{A} \cup \overline{B}$, I attempted something myself. What I did is nothing like the proof given in the text, and might be needlessly complicated, but I wanted to check that it still works. I don't always trust my judgment that the mathematical logic is correct, so sometimes an extra set of eyes is useful.

First, perhaps I should define $\overline{A}$. It is the set of adherence points of the set $A\subset X$ meaning it is the set of all points $x\in X$ such that any neighborhood of $x$ also contains points of $A$.

Now, here is my proof:

Suppose $x\notin \overline{A}\cup \overline{B}$. Then, $x\notin \overline{A}$ and $x\notin \overline{B}$.

$x\notin \overline{A}$ implies that $\exists$ neighborhood $U(x)$ such that $x \in U(x)$ and $U(x) \cap A =\emptyset$, which implies that $x\in A^{c}$.

Likewise, $x\notin \overline{B}$ means that $\exists$ a neighborhood $W(x)$ such that $x\in W(x)$ and $W(x)\cap B = \emptyset$. This implies that $x\in B^{c}$.

So, $x\in A^{c}$ and $x\in B^{c}$, which means that $x \in A^{c}\cap B^{c}\subset\overline{A^{c}\cap B^{c}}\, \implies \, x\notin \overline{A\cup B}$

So, I showed that $x\notin \overline{A}\cup \overline{B}\, \implies \, x\notin \overline{A \cup B}$, which is the contrapositive of $x\in \overline{A \cup B}\, \implies \, x\in \overline{A} \cup \overline{B}$, at least I think it is. I've experienced some setbacks as of late that have caused me to lose a little bit of confidence in my judgment when it comes to mathematical logic.

Thanks in advance.

Answer 1

Your proof sets itself up well, but then ends confusedly. In particular, you first note:

If $x\not \in \bar A$, then there is an open set $U$ containing $x$ disjoint from $A$.

and similarly

If $x\not \in \bar B$, then there is an open set $W$ containing $x$ disjoint from $B$.

But then you do not use these premises and merely claim that since $x$ is in $\overline{A^C\cap B^C}$ it must not be in $\overline{A\cup B}$. This is not a valid argument - these two sets intersect exactly on the boundary of $A\cup B$.

Rather, to conclude that $x\not \in \overline{A\cup B}$, you need to show there is an open set $V$ containing $x$ disjoint from $A\cup B$. This is easy given what you showed earlier: Just take $V=U\cap W$.