Vtyjkyuk

For any two sets $A$ and $B$ in a space $X$, $$textcl(A cup B)=textcl(A) cup textcl(B)$$ where $textcl(Y)$ denote the closure of $Y$

This question and its answer are already available in this site. Why I am posting here is "to check my proof". Here's My try:

$$x in textcl(A cup B) Leftrightarrow exists ;r>0 ;textsuch that;B_r(x) cap (A cup B) neq emptyset$$

$$Leftrightarrow (B_r(x) cap A)cup (B_r(x) cap B) neq emptyset$$

$$Leftrightarrow (B_r(x) cap A) neq emptyset ;textor;(B_r(x) cap B) neq emptyset$$

$$Leftrightarrow x in textcl(A); textor; x in textcl(B)$$

$$Leftrightarrow x in textcl(A) cup textcl(B)$$

Is this right?

I take issue with your first equivalence. Take the open ball of radius $1$ and center $0$. Then for a point $p$, a ball $B_r(p)$ intersects $B_1(0)$ if you take $r = mathrmdist(p,0)$. Is the closure of $B_1(0)$ the full space $mathbb R^n$ ?

The equivalence you should have is:

$$
xin mathrmcl(x) Leftrightarrow forall r > 0: B_r(x) cap(A cup B) ne varnothing
$$

But you have an issue in the sense that for a given $x$, $B_r(x)$ will intersect either with $A$ or $B$ and so far that depends on $r$. But you must have either $A$ or $B$ that intersects with $B_r(x)$ for all $r > 0$, not some of them depending on $r$, so you have additional steps to do.

Note that if there is a radius $r_1$ such that $B_r_1(x)$ intersects $A$ but not $B$, for any $0 < r_2 < r_1$, $B_r_2(x)$ cannot intersect $B$, else you contradict the conditions on $r_1$. Moreover, for $r_3 > r_1$, $B_r_3(x)$ must intersect $A$ since $B_r_1(x) subset B_r_3(x)$. Thus, you can see that for a given $x$, if there is an $r_1$ such that $B_r_1(x)$ does not intersect $B$, then for all $r>0$, $B_r(x)$ intersects $A$.

This answer shows that the equality is valid in every topological space.

From $Asubseteq Acup B$ it follows directly that $mathsfcl(A)subseteqmathsfcl(Acup B)$.

Similarly we find $mathsfcl(B)subseteqmathsfcl(Acup B)$ and conclude that:

$$mathsfcl(A)cupmathsfcl(B)subseteqmathsfcl(Acup B)$$

If $xnotinmathsfcl(A)cupmathsfcl(B)$ then open sets $U,V$ exist that contain $x$ as element and have an empty intersection with $A$ and $B$ respectively.

Then $Ucap V$ is an open set that contains $x$ as element and has an empty intersection with $Acup B$. Proved is then that $xnotinmathsfcl(A)cupmathsfcl(B)$ implies that $xnotinmathsfcl(Acup B)$. This justifies the conclusion that also:$$mathsfcl(Acup B)subseteqmathsfcl(A)cupmathsfcl(B)$$and we conclude that:$$mathsfcl(Acup B)=mathsfcl(A)cupmathsfcl(B)$$

More shortly the second part follows also from the following reasoning: $mathsfcl(Acup B)$ is the smallest closed set that contains $Acup B$ as a subset. The sets $mathsfcl(A)$ and $mathsfcl(B)$ are both closed, and consequently their union is closed. Further this union contains $Acup B$ as a subset so we are allowed to conclude that $mathsfcl(Acup B)$ is a subset of this union.