closure of finite unions

Solution 1:

It is true.

$A \subset A \cup B $ and $B \subset A \cup B$ implies $\textrm{cl}(A) \subset \textrm{cl}(A \cup B)$ and $ \textrm{cl}(B) \subset \textrm{cl}(A \cup B)$. Hence $ \textrm{cl}(A) \cup \textrm{cl}(B) \subset \textrm{cl}(A \cup B)$.

$\textrm{cl}(A) \cup \textrm{cl}(B)$ is closed. (since the two component sets are closed) Also, we know that $A\subset \textrm{cl}(A) $ and $ B\subset \textrm{cl}(B).$ Hence, $(A \cup B) \subset \textrm{cl}(A) \cup \textrm{cl}(B) $ and it follows that $\textrm{cl}(A \cup B) \subset \textrm{cl}(A) \cup \textrm{cl}(B) $

So the result is true for any two sets. We can extend this result to any finite number of sets using induction. Thus, the result you wanted to prove holds.

Solution 2:

It is. The closure of a set is the smallest closed superset (a good exercise, if you've not encountered that result before), and a union of finitely many closed sets is closed. From that, proving double inclusion is fairly straightforward.

Suppose $A_1,...,A_n$ are arbitrary subsets of the topological space $X$. For any $1\leq j\leq n$, we have $$A_j\subseteq\bigcup_{m=1}^nA_m\subseteq\mathrm{cl}\left(\bigcup_{m=1}^nA_m\right),$$ so since $\mathrm{cl}(A_j)$ is the smallest closed superset of $A_j$, then $$\mathrm{cl}(A_j)\subseteq\mathrm{cl}\left(\bigcup_{m=1}^nA_m\right),$$ and since this holds for all $1\leq j\leq n$, then $$\bigcup_{m=1}^n\mathrm{cl}(A_m)\subseteq\mathrm{cl}\left(\bigcup_{m=1}^nA_m\right).$$

Remark: The above portion of the containment also works with infinitely many sets, meaning $\bigcup_{j\in J}\mathrm{cl}(A_j)\subseteq\mathrm{cl}\left(\bigcup_{j\in J}A_j\right)$ for any indexed set $\{A_j\}_{j\in J}$ of subsets of $X$.

On the other hand, we also have for each $1\leq j\leq n$ that $$A_j\subseteq\mathrm{cl}(A_j)\subseteq\bigcup_{m=1}^n\mathrm{cl}(A_m),$$ so since that holds for all $1\leq j\leq n$, we have $$\bigcup_{m=1}^nA_m\subseteq\bigcup_{m=1}^n\mathrm{cl}(A_m).$$ Now $\mathrm{cl}\left(\bigcup_{m=1}^nA_m\right)$ is the smallest closed superset of $\bigcup_{m=1}^nA_m$, so since a union of finitely many closed sets is closed--meaning in particular that $\bigcup_{m=1}^n\mathrm{cl}(A_m)$ is closed--we have that $$\mathrm{cl}\left(\bigcup_{m=1}^nA_m\right)\subseteq\bigcup_{m=1}^n\mathrm{cl}(A_m),$$ and so $$\bigcup_{m=1}^n\mathrm{cl}(A_m)=\mathrm{cl}\left(\bigcup_{m=1}^nA_m\right)$$ by double inclusion.

Remark: The second half of the inclusion might not hold for infinitely-many sets (it does sometimes). Consider, for each positive integer $n$, the real interval $$A_n:=\left(\frac{1}{n+1},\frac{1}{n}\right),$$ with $X:=\Bbb R$ in the standard topology. Clearly, $\mathrm{cl}(A_n)=\left[\frac{1}{n+1}\frac 1 n\right]$, from which we can show that $$\bigcup_{n=1}^\infty\mathrm{cl}(A_n)=(0,1]$$...but that isn't even closed, so can't possibly be the same thing as $\mathrm{cl}\left(\bigcup_{n=1}^\infty A_n\right)$. Fortunately, we do know by the first part that $$\bigcup_{m=1}^\infty A_n\subseteq (0,1]\subseteq\mathrm{cl}\left(\bigcup_{n=1}^\infty A_n\right),$$ and so the smallest closed superset of $\bigcup_{n=1}^\infty A_n$ is also a superset of $(0,1]$. Well, what is the smallest closed superset of $(0,1]$? That has to be $[0,1]$, so $$\bigcup_{m=1}^\infty \mathrm{cl}(A_n)=(0,1]\subset [0,1]=\mathrm{cl}\left(\bigcup_{n=1}^\infty A_n\right)$$ in this case. There are other ways to show that $\mathrm{cl}\left(\bigcup_{m=1}^\infty A_n\right)=[0,1]$, but this is probably the most direct.