Who's right regarding this proof of a property of the boundary of a union? The question is to prove $$\partial(A\cup B)\subseteq\partial A\cup \partial B.$$
My proof is, let $z$ be in $\partial (A \cup B)$. Then for any $\epsilon > 0$, the definition of boundary says $D_\epsilon (z)\cap (A \cup B) \neq \emptyset$ and $D_\epsilon (z)\cap (A \cup B)^c \neq \emptyset$.
This implies two statements,

*

*$\;\;D_\epsilon (z)\cap A \neq \emptyset$ or $D_\epsilon (z)\cap B \neq \emptyset$

*$\;\;D_\epsilon (z)\cap A^c \neq \emptyset$ and $D_\epsilon (z)\cap B^c \neq \emptyset$
which implies either $z \in \partial A$ or $z\in \partial B$, again from the definition of boundary.
My professor marked my proof incorrect by saying, $$\forall \epsilon >0 (x\:or\:y)$$ does not imply $$(\forall \epsilon >0\:x)\: or \: (\forall \epsilon >0\:y)$$
I agree with this statement, but I don't understand how that applies to my proof.
 A: This is pretty neat.
Suppose we have two values of $\epsilon$, let's call them $\epsilon_1$ and $\epsilon_2$. To save some time I will write $D_1$ instead of $D_{\epsilon_1}$ and similarly for $D_2$.
Now by your proof we have that at least one of the following two statements hold:
a) $D_1(z) \cap A \neq \emptyset$ and $D_1(z) \cap A^c \neq \emptyset$
b) $D_1(z) \cap B \neq \emptyset$ and $D_1(z) \cap B^c \neq \emptyset$
Also by your proof at least one of the following two statements hold:
c) $D_2(z) \cap A \neq \emptyset$ and $D_2(z) \cap A^c \neq \emptyset$
d) $D_2(z) \cap B \neq \emptyset$ and $D_2(z) \cap B^c \neq \emptyset$
Now an all knowing oracle happens to be visiting town that day so you grab the opportunity to ask which one it is. The oracle tells you:
'a) and d) are true and b) and c) are false for this particular values of $A, B$ and $z$.'
Okay, so far so good. All this is perfectly consistent with your proof. But now it turns out your professor overheard the conversation and says:
'Aha! You claimed that $z$ was in the boundary of $A$ and I was inclined to believe you, but now it seems you lied to me! $z$ is not in the boundary of $A$ because the oracle just tells me statement c) is false! So there is at least one $\epsilon$ (namely $\epsilon_2$) for which we do not have that $D_\epsilon(z)$ intersects both $A$ and $A^c$.'
'Wait a moment, wait a moment', you reply. 'I never claimed that $z$ was in the boundary of $A$, just that it was in the boundary of $A$ OR in the boundary of $B$.' 'Alright', your professor replies. 'So you agree that it is in the boundary of $B$ then? This seems a bit odd given the falsehood of statement b... Since when does $\epsilon_1$ not belong to "all $\epsilon$"?'
I hope that this slightly dramatized version of your conversation shows what the professor was trying to convey.
I'd say your proof is not wrong per se, but it is incomplete. There must be an extra argument that the situation above (with $a$ and $d$ true, but $b$ and $c$ false) cannot occur in reality.
A: My proof: In any topological space we have $\overline {A\cup B}=\overline A\cup \overline B$. And we have $\partial D=\overline D \cap \overline {D^c}.$ And we also have $$\overline {(A\cup B)^c}\subset \overline {A^c}\cap\overline {B^c}$$ because $$\overline {(A\cup B)^c}=\overline {A^c\cap B^c}\subset \overline {A^c}$$ and $$\overline {(A\cup B)^c}=\overline {A^c\cap B^c}\subset \overline {B^c}.$$ Therefore $$\partial (A\cup B)=(\overline {A\cup B})\cap \overline {(A\cup B)^c}\subset$$ $$\subset (\overline {A\cup B}) \cap \overline {A^c}\cap\overline {B^c}=$$ $$=(\overline A \cup \overline B)\cap \overline {A^c}\cap\overline {B^c}=$$ $$=(\overline A \cap \overline {A^c}\cap\overline {B^c})\cup (\overline B\cap \overline {A^c}\cap\overline {B^c})\subset$$ $$\subset  (\overline A \cap \overline {A^c} )\cup (\overline B \cap \overline {B^c})=$$ $$=(\partial A)\cup (\partial B).$$
