# Topology closure

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)?

• Yes, they are equivalent. – Kavi Rama Murthy Jul 9 at 0:29
• @KaviRamaMurthy why can’t it be: $\forall U$ $( U \in \tau$ and $x\in U \textbf{and} U \cap A$ is not empty)$? – topologicalmagician Jul 9 at 0:32 • @topologicalmagician Because that would assert that, for every (not necessarily open) subset of$X$, that$U \in \tau$(i.e. every subset is open) and$x \in U$(every subset contains$x$) and$U \cap A \neq \emptyset$(every subset intersects$A$). This is never true, as it doesn't exclude$U = \emptyset\$. – Theo Bendit Jul 9 at 1:19

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

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

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.