Closure of the union = Union of closures

Question

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

Answer 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 contains both $A$ 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 contains $A \cup B$, so that any closed set which contains $A\cup B$ also contains $\text{cl}(A \cup B)$. Therefore $\text{cl}(A \cup B) \subset \text{cl}(A) \cup \text{cl}(B)$.

Answer 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.

Answer 3

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 $x \in B$, $\exists p \in A_1 \vee \exists k \in A_2: p \in B \vee k \in B$, 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 $\overline{(A_1 \cup A_2)} = \overline A_1 \cup \overline A_2$

Answer 4

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 $x \in Cl(A)\cup Cl(B)$.

Answer 5

We have to proof that $\overline{A \cup B} = \overline{A} \cup \overline{B}$

$(\subset)$

\begin{align} x \notin \overline{A \cup B} & \iff \exists V_x (\text{neighborhood of x}) : V_x \cap (A \cup B) = \varnothing\\ &\iff(V_x \cap A) \cup (V_x \cap B) = \varnothing\\ &\implies V_x \cap A= \varnothing \text{ and } V_x \cap B = \varnothing\\ &\iff x \notin \overline{A} \text{ and } x \notin \overline{A}\\ &\iff x \notin \overline{A} \cup \overline{B} \end{align}

$(\supset)$ \begin{align} x \in \overline{A} \cup \overline{B} & \iff x \in \overline{A} \text{ or } x \in \overline{B}\\ &\iff \forall V_x : V_x \cap A \ne \varnothing \text{ or } V_x \cap B \ne \varnothing\\ &\implies (V_x \cap A) \cup (V_x \cap B) \ne \varnothing\\ &\iff V_x \cap (A \cup B) \neq \varnothing\\ &\iff x \in \overline{A \cup B} \end{align}

Answer 6

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

Answer 7

I got stuck at the same point. I was able to use contraposition to make things easier - I believe this works:

We want to show that $x\in (H\cup K)' \Rightarrow x \in H' \cup K'$ (and thus $x\in \overline{H} \cup \overline{K}$ )

By way of contraposition, suppose $x \notin H' \cup K'$

Thus there exists an open $D$ containing $x$ which contains no points of either $H$ or of $K$ distinct from $x$. In other words, $D$ contains no points of $H\cup K$ distinct from $x$.

Thus $x \notin(H\cup K)'$

Answer 8

This is just for fun. The “abstract nonsense” can be converted into a more concrete proof.

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$\newcommand{\P}{\mathscr{P}}\newcommand{\C}{\mathscr{C}}\newcommand{\cl}{\operatorname{cl}}$Fix a topological space $X$ and let $\P(X),\C(X)$ be viewed as posetal categories where $\P$ denotes power set and $\C$ denotes the collection of all closed sets.

Then there is a forgetful $\iota:\C(X)\hookrightarrow\P(X)$ and the closure operator $\cl:\P(X)\to\C(X)$ is its left adjoint. That is, $\cl$ is the reflector for the reflective subcategory $\C(X)\subseteq\P(X)$.

In $\P(X)$, (finite) unions are viewable as colimits (namely, a union will be the coproduct of the unionands) and in $\C(X)$ we have that colimits of finite diagrams are also given by the union operation.

Then by the theorem: “left adjoints are cocontinuous”, and the observation that isomorphism is equality in these categories: $$\cl\left(\bigcup_{j=1}^n A_j\right)=\bigcup_{j=1}^n\cl(A_j)$$

Follows for all $A_1,\cdots,A_j\in\P(X)$.

The same proof, indeed the dual proof, shows that the interior of a finite intersection is the finite intersection of the interiors.