# 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

• Choose some limit point of LHS and observe that it belong to the RHS.
– user173262
Commented Oct 26, 2016 at 16:06

(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 $$\text{cl}(A \cup B) \subset \text{cl}(A) \cup \text{cl}(B)$$.

• But how come $x$ \in cl(A)? I have a neigbourhood of $x$ only intersecting at B\A so this neigbourhood does not intersect A so it shouldn't be a limitpoint of A right? Commented Oct 26, 2016 at 16:36
• @TheGeometer Indeed, I've got confused! Define your two open sets to be $U_1$ and $U_2$. Then $U_1 \cap U_2$ is a third open set containing $x$, that neither intersect $A$ or $B$. So $x \not\in \text{cl}(A \cup B)$ ;). So it's not a counterexample. Commented Oct 26, 2016 at 16:44
• Exactly! Thanks man I got so paranoid with that problem:P Commented Oct 26, 2016 at 16:45
• Is the proof for the first part valid for an arbitrary collection of sets? Commented Nov 2, 2019 at 12:00
• @H.R. No, consider $\{x \in (0,1)\}$. Then, $(0,1)=\cup_{x} cl\{x\} \subseteq cl(\cup_x \{x\})=[0,1]$. If you post as a separate question; I will poste this as an answer :)
– ABIM
Commented Dec 2, 2019 at 15:04

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.

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 $$\overline{(A_1 \cup A_2)} = \overline A_1 \cup \overline A_2$$

• Wonderful answer! Maybe I humbly suggest though, it is good practice to use brackets when writing logic propositions. E.g. writing $(p \in A_1) \lor (p \in A_2)$ will be clearer than writing $p \in A_1 \lor p \in A_2$ Commented Feb 9, 2023 at 17:30
• You are absolutely right. I wrote this answer before I started studying math in college, and now, after some years, I totally agree with you. @Newton Commented Feb 9, 2023 at 20:57

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

• This is because if $y\in Cl(A)$ iff every open set containing $y$ intersects $A$. Commented Oct 26, 2016 at 16:10
• I think I have a confusion with the definition. How I know it is that $x$ is a limit point of a subset $S$ of a topological space if every neigbourhood of $x$ intersects with $S$. So in our case if $x$ has a neigbourhood only intersecting $A\B$ and another only intersecting $B\A$ it should be a limit point only of $A\cupB$ and not of A or B.Am I using wrond definition or saying something wrong? Commented Oct 26, 2016 at 16:19
• The $Cl(A)$ is the limit points unioned with the set A. So $Cl(A)=A \cup A'$. I believe this is the confusion, though I am slightly confused on what your counterexample is saying. Commented Oct 26, 2016 at 16:57

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}

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

• ok but what is wrong wit my counterexample? Commented Oct 26, 2016 at 16:04

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

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

$$\newcommand{\P}{\mathscr{P}}\newcommand{\C}{\mathscr{C}}\newcommand{\cl}{\operatorname{cl}}$$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.