Let $(X,d)$ be a metric space and let $A \subseteq X$.
Prove that $A$ is dense if and only if any non-empty open set in $X$ has an intersection with $A$

In my proof $\text{Cl}(A)$ denotes the closure of $A$.

I need to prove, equivalently, that $\forall O$ non-empty open subset of $X$: $$\text{Cl}(A)=X\Longleftrightarrow O\cap A\neq \emptyset$$

I feel my attempt has some logical errors, and I kindly request criticism and insight.
If my attempt is entirely wrong, please help me with alternative ways.
Thank you very much.

Here is my attempt:

PART 1 ($\Longrightarrow$):

We have $\text{Cl}(A) = X$, therefore by definition of the closure: $$\forall a\in X:\forall N\in N(a), N\cap A\neq\emptyset$$ Every neighborhood of every point of $X$ intersects $A$.

Therefore one could add, every open neighborhood of every point of $X$ intersects $A$.

Let $N_O(a)\subseteq N(a)$ be the set of open neighborhoods of $A$, then by definition: $$\forall a\in X: \forall O\in N_O(a),O\cap A\neq\emptyset$$ Therefore, every non-empty open neighborhood of every point of $X$ intersects $A$.

In other words, every non-empty open subset of $X$ intersects $A$.

PART 2 ($\Longleftarrow$):

We have $\forall O \subseteq X$ that $O \cap A \neq \emptyset$.
($O$ is non-empty and open)

Let $a \in O$.

Assume $\text{Cl}(A) \neq X$, therefore $\text{Cl}(A)^\text{C} \neq \emptyset$.

By negating the definition of the closure of $A$, we get $\forall a\in \text{Cl}(A)^\text{C}$: $$\exists N \in N(a): N\cap A = \emptyset$$ $N$ is a neighborhood for $a$, therefore $\exists O$ open: $a \in O \subseteq N$.

Therefore, $O \cap A = \emptyset$ but $O \cap A \neq \emptyset$.

Contradiction, therefore $\text{Cl}(A) = X$.

  • $\begingroup$ The definition I am using is: $a \in \text{Cl(A)} \Longleftrightarrow \forall N \in N(a), N \cap A \neq \emptyset$. $\endgroup$ – ex.nihil Feb 22 '18 at 17:17
  • $\begingroup$ Thank you for the rigid tips, I have this nasty habit of writing proofs as if I am programming, hence why the redundancy sometimes. $\endgroup$ – ex.nihil Feb 22 '18 at 17:37

Be careful with your statements. You don't have to prove that

for every non empty open set $O$ of $X$, $\operatorname{Cl}(A)=X$ if and only if $O\cap A\ne\emptyset$

but rather

$\operatorname{Cl}(A)=X$ if and only if, for every non empty open set $O$ of $X$, $O\cap A\ne\emptyset$

which is quite a different statement.

Your part 1 is correct, but full of unnecessary bits. What you have to prove is that, assuming $\operatorname{Cl}(A)=X$, that every nonempty open set $O$ intersects $A$. This follows from the fact that, considering $x\in O$, $O$ is an open neighborhood of $x$; since $x\in\operatorname{Cl}(A)$, we conclude that $O\cap A\ne\emptyset$.

Part 2 is correct, although lengthier than needed. Suppose $x\notin\operatorname{Cl}(A)$. Then, by definition of closure, there exists an open neighborhood $O$ of $x$ such that $O\cap A=\emptyset$.

Note. No contradiction is necessary in part 2, we're proving the contrapositive, that is, if $\operatorname{Cl}(A)\ne X$, then there exists a nonempty open set $O$ such that $O\cap A=\emptyset$.

  • $\begingroup$ Very helpful, thank you. Excuse my ignorance, but precisely how are the statements different? $\endgroup$ – ex.nihil Feb 22 '18 at 17:50

The logic of your attempt seems right but I am sure your argument can be shortened to make it clearer. For example:

If $\mathrm{Cl}(A)=X$, then for any non-empty open $U\subset X$, just pick an $x\in U$ and $U\cap A\neq\varnothing$ follows.

Conversely if $U\cap A\neq\varnothing$ for any non-empty open $U\subset X$, then each $x\in X$ lies in $\mathrm{Cl}(A)$ since each neighborhood of $x$ intersects with $A$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.