Spotting error in proof for $(A \cup B)^\circ = A^\circ \cup B^\circ$ I wanted to show that $(A \cap B)^\circ = A^\circ \cap B^\circ$. I came up with a proof, but I don't know why it works. 
$$y \in (A \cap B)^\circ$$ 
$$\Leftrightarrow \exists \epsilon > 0: U_{\epsilon} (y) \subset A \cap B$$ 
$$\Leftrightarrow \exists \epsilon > 0: x \in U_{\epsilon} (y) \Rightarrow x\in A \land x\in B$$
$$\Leftrightarrow \exists \epsilon > 0: [x \in U_{\epsilon} (y) \Rightarrow x\in A] \land [ x \in U_{\epsilon} (y) \Rightarrow x\in B]$$
$$\Leftrightarrow [\exists \epsilon > 0: x \in U_{\epsilon} (y) \Rightarrow x\in A] \land [\exists \epsilon > 0: x \in U_{\epsilon} (y) \Rightarrow x\in B]$$
$$\Leftrightarrow y \in A^\circ \land y \in B^\circ$$
$$\Leftrightarrow y \in A^\circ \cap B^\circ$$
So, here is why I don't know why it works. Because as it seems, I could replace all of the $\cap$ by a $\cup$ and all of the $\land$ by a $\lor$ and all of the equivalences should still work, which I know they shouldn't. So which equivalence fails for $\lor$ and why?
 A: I'll the intersection proof in my own words:
Suppose that $x \in (A\cap B)^\circ$. This means there is some $r>0$ such 
that $B(x,r) \subseteq A \cap B$. But as $A \cap B \subseteq A$ we see that $B(x,r) \subseteq A$ and thus $x \in A^\circ$. As we also have $A \cap B \subseteq B$, $B(x,r) \subseteq B$ and thus by definition $x \in B^\circ$. As we have both $x \in A^\circ$ and $x \in B^\circ$, $x \in A^\circ \cap B^\circ$. This shows that $(A \cap B)^\circ \subseteq A^\circ \cap B^\circ$.
Now suppose that $x \in A^\circ \cap B^\circ$. Then in particular, $x \in A^\circ$, so there is some $r_1 > 0$ such that $B(x,r_1) \subseteq A$.
Also, $x \in B^\circ$ so there is some $r_2 > 0$ such that $B(x,r_2) \subseteq B$. Now define $r=\min(r_1, r_2) > 0$. Then $r \le r_1$ so $B(x,r) \subseteq B(x,r_1)$ and $r \le r_2$ so $B(x,r) \subseteq B(x,r_2)$. Hence $B(x,r) \subseteq B(x, r_1) \cap B(x, r_2) \subseteq A \cap B$, which shows that $x \in (A \cap B)^\circ$, and as $x$ was arbitrary, $A^\circ \cap B^\circ \subseteq (A \cap B)^\circ$, showing the final inclusion for the equality.
For the union things go wrong:
Suppose that $x \in (A \cup B)^\circ$. Then there is some $r>0$ such that $B(x,r) \subseteq A \cup B$. So I know that all $y$ with $d(x,y) < r$ are in $A$ or in $B$ (or both), but the set they're in can differ for each point in $B(x,r)$, so I cannot say that $B(x,r) \subseteq A$ or $B(x,r) \subseteq B$ always. So I cannot make progress to show that $x \in A^\circ$ or $x \in B^\circ$. 
The reverse inclusion does proceed smoothly: suppose $x \in A^\circ \cup B^\circ$. Then $x \in A^\circ$ or $x \in B^\circ$. E.g. $x \in A^\circ$ holds, and thus we have $r>0$ such that $B(x,r) \subseteq A$, but $A \subseteq A \cup B$ so that trivially $B(x,r) \subseteq (A \cup B)$. A similar conclusion is reached when $x \in B^\circ$ of course. So then $x \in (A \cup B)^\circ$ and thus $A^\circ \cup B^\circ \subseteq (A \cup B)^\circ$.
An example where we see the complete failure of the other implication: $A = \mathbb{Q}$, $ B = \mathbb{R}\setminus \mathbb{Q}$ in the reals (usual topology), where $A^\circ  =B^\circ  = \emptyset$ but $(A \cup B)^\circ  = \mathbb{R}^\circ = \mathbb{R}$.
A: Refer to the figure below:
$\hspace{7cm}$
Note that $y \in (A \cup B)^\circ$, however $y \notin A^\circ \vee y \notin B^\circ$. Hence: $(A \cup B)^\circ \ne A^\circ \cup B^\circ$.
The correct formula must be: $A^\circ \cup B^\circ \subseteq  (A \cup B)^\circ$.
A: Fourth equivalence fails if we replace every $\cap$ and $\land$ to $\cup$ and $\lor$. It will assert the existence of $\epsilon>0$ such that $U_\epsilon(y)\subset A\cup B$ implies there is $\epsilon>0$ with $U_\epsilon(y)\subset A$ or $U_\epsilon(y)\subset B$ and vice versa. It does not actually hold, however. (For example, consier $A$ be the set of rationals and $B=\mathbb{R-Q}$.)
A: $x \in U_\epsilon(y) \implies (x \in A \vee x \in B)$ does not imply
$(x \in U_\epsilon(y) \implies x \in A) \vee (x \in U_\epsilon(y) \implies x \in B)$.
