Topology closure

Question

Let $(X, \tau)$ be a topological space. Let $A\subset X$

My professor defined the closure of $A$ as the set of points in $X$ so that $\forall$ $U$(open) containing $x$, $U \cap A$ is non-empty.

The question I have is which of the following ones are logically equivalent to the one the Professor gave?

$\forall$ $U$(open) containing $x$, $U \cap A$ is non-empty.

(1) $\forall U$ ($ U \in \tau$ and $ x\in U \implies U \cap A $ is not empty)?

Or

(2) $\forall U$ ($ U \in \tau$ and $ x\in U \textbf{ and } U \cap A $ is not empty)?

Or

(3) $\forall U$ ($ U \in \tau$ $\implies$ $ x\in U \textbf{ and } U \cap A $ is not empty)?

Please explain why.

Answer 1

Only 1 is correct: when we quantify over all open sets (or equivalently over all $U \in \tau$)) we know the implication if $x \in U$ then $U$ must intersect $A$. We do not know that all open sets contain $x$; this will be exceedingly rare or even false as $\emptyset$ is open in any $X$..

So correct is

$$\forall U \in \tau: (x \in U \implies U \cap A \neq \emptyset)$$

Answer 2

1 is correct.

2 implies that all open open sets contain $x$.
That is false. For example $\Bbb R - \{x\}$ is open and excludes $x$.

3 implies that if $U$ is an open set, then $x \in U$
which again is false. $U=\emptyset$, for example.