Closure of the union = Union of closures

I have seen that $\text{cl}(A\cup B)=\text{cl}(A)\cup \text{cl}(B)$. However I don't see why this is true. I can see the the right to left inclusion, but I can't see the inclusion from left to right. Say that I have an element $x$ contained only in two open sets one that intersects $A\cup B$ only in $A\setminus B$ and and another that intersects only in $B\setminus A$ isn't this a contradiction?

Edit: I have seen the proof but I still can't understand what is wrong with the counterexaple above

Thanks in advance

Solution 1:

(1) ($\supset$) :: \begin{align*} A \subset A \cup B \implies \text{cl}(A) \subset \text{cl}(A \cup B) \\B \subset A \cup B \implies \text{cl}(B) \subset \text{cl}(A \cup B) \end{align*} therefore yielding that $\text{cl}(A) \cup \text{cl}(B) \subset\text{cl}(A \cup B)$

(2) ($\subset$) ::

The subset $\text{cl}(A) \cup \text{cl}(B)$ is closed and both contains $A$ and $B$, therefore $A \cup B \subset \text{cl}(A) \cup \text{cl}(B)$. $\text{cl}(A \cup B)$ is defined to be smallest closed set which contained $A \cup B$, so that any closed set which contained $A\cup B$ also contains $\text{cl}(A \cup B)$. Therefore $\text{cl}(A \cup B) \subset \text{cl}(A) \cup \text{cl}(B)$.

Solution 2:

In your proposed counterexample, you've forgotten that open sets are closed under finite intersection.

So if $x$ has a neighborhood that only meets $A\cup B$ in $B\setminus A$ and a neighborhood that only meets $A\cup B$ in $A\setminus B$, then the intersection of these neighborhoods doesn't meet $A\cup B$ at all.

Solution 3:

Let $x\in Cl(A\cup B)$ then every open set containing $x$ intersects $A\cup B$. Thus for $x \in U_\alpha$ where $U_\alpha$ is open in $X$. If $U_\alpha$ intersects $A$, then $x \in Cl(A)$ else $x \in Cl(B)$ either way $x \in Cl(A)\cup Cl(B)$.

Solution 4:

Infinite point sequence from (A ∪ B) contains an infinite subsequence from A or contains an infinite subsequence from B

Solution 5:

Let's use the following definition of closure:

Let $A$ be a subset of $(X,\tau)$. Then the set $A \cup A'$, consisting of the set $A$ and all it's limit points it's called the closure of $A$ and is denoted by $\overline A$.

So, we want to prove that $\overline{(A_1 \cup A_2)} = \overline A_1 \cup \overline A_2$.

Part 1

Let's start by proving that $\overline{(A_1 \cup A_2)} \subseteq \overline A_1 \cup \overline A_2$.

Let $x \in \overline{(A_1 \cup A_2)}$, Then we have that $x \in (A_1 \cup A_2) \cup (A_1 \cup A_2)'$ (this is the definition of closure).

If $x \in (A_1 \cup A_2)$, then, because $\overline A_1 \cup \overline A_2 = (A_1 \cup A_2) \cup (A_1' \cup A_2') $, we have that $x \in \overline A_1 \cup \overline A_2$.

If $x \in (A_1 \cup A_2)'$, then we have that $x$ is a limit point of the set $A_1 \cup A_2$. By the definition of limit point this means that, for every open set $B \in \tau$ such that $x \in B$, $\exists p \in A_1 \cup A_2: p \in B$ and $p \neq x$. because $p \in A_1 \cup A_2$ we have that $p \in A_1 \vee p \in A_2$ whitch is the same as saying that $x$ is a limit point of $A_1$ $\vee$ $x$ is a limit point of $A_2$. Thus $x \in (A_1' \cup A_2') \to x \in \overline A_1 \cup \overline A_2$.

This means that $\overline{(A_1 \cup A_2)} \subseteq \overline A_1 \cup \overline A_2$.

Part 2

Now let's prove that $\overline{(A_1 \cup A_2)} \supseteq \overline A_1 \cup \overline A_2$

Let $x \in \overline A_1 \cup \overline A_2$. Then $x \in (A_1 \cup A_2) \cup (A_1' \cup A_2')$.

If $x \in (A_1 \cup A_2)$, it's trivial that $x \in (A_1 \cup A_2) \cup (A_1 \cup A_2)' = \overline(A_1 \cup A_2)$

If $x \in (A_1' \cup A_2')$, then $x$ is a limit point of $A_1$ or a limit point of $A_2$. This tells us that, for every $B \in \tau$ such that $x \in B$, $\exists p \in A_1 \vee k \in A_2: p \in B \vee k \in B$, such that both $p$ and $k$ are different from $x$. But $p \in A_1 \subseteq A_1 \cup A_2$, and $k \in A_2 \subseteq A_1 \cup A_2$. So we have that $p,k \in A_1 \cup A_2$. So $x$ is also a limit point of $A_1 \cup A_2 \to x \in \overline{(A_1 \cup A_2)}$. This implies that: $\overline{(A_1 \cup A_2)} \supseteq \overline A_1 \cup \overline A_2$

Part 1 + Part 2 (Conclusion)

From part 1 we deduced that $\overline{(A_1 \cup A_2)} \subseteq \overline A_1 \cup \overline A_2$, and from part 2 $\overline{(A_1 \cup A_2)} \supseteq \overline A_1 \cup \overline A_2$. This leads us to the conclution that $\overline{(A_1 \cup A_2)} = \overline A_1 \cup \overline A_2$