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Can someone check if this proof is correct, please? Im not sure if it is enough formal or complete. Thank you.

Let $M$ be a metric space. Prove or disprove that finite intersection of open dense subsets of $M$ are open and dense in $M$.

Let $A, B$ be open dense subsets in $M$. Hence for every $x\in M$ and any $\epsilon>0$ there is some $a\in A$ such that $a\in\Bbb B(x,\epsilon)$.

Now: because $A$ is open then all it points are interior points, that is, for every $a\in A$ there is a $\delta>0$ such that $\Bbb B(a,\delta)\subseteq A$. Then is clear that for enough small $\delta>0$ we have that

$$a\in\Bbb B(x,\epsilon)\implies\Bbb B (a,\delta)\subset\Bbb B(x,\epsilon)$$

In particular this is also true for any $b\in B$, and because $B$ is open then

$$a\in\Bbb B(b,\epsilon)\implies\Bbb B(a,\delta)\subset\Bbb B(b,\epsilon)\subset B$$

for some enough small $\delta>0$. Then for any neighborhood of $x\in M$ we have that

$$\Bbb B(a,\delta)\subset\Bbb B(b,\epsilon)\subset\Bbb B (x,\eta)\tag{1}$$

or

$$\Bbb B(b',\rho)\subset\Bbb B(a,\delta)\subset\Bbb B (x,\eta)\tag{2}$$

for some $b'\in B$, because $B$ is open and dense in $M$. Hence the intersection of $A$ and $B$ is clearly dense in $M$, that is without lose of generality from (1) we have that

$$\Bbb B(a,\delta)\cap\Bbb B(b,\epsilon)=\Bbb B(a,\delta)$$

And $A\cap B$ is open by the axiomatic definition of a topological space, that is, by definition finite intersection of open sets are open.$\Box$

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    IMHO you're overcomplicating things.. – 2017-02-05
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    @HennoBrandsma yes I noticed... – 2017-02-05

2 Answers 2

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This follows from general facts (nothing specific for metrics):

  1. The finite intersection of open sets is always open, by axiom 2 of a topology (this does not need a proof).

  2. If $O$ is open and $D$ is dense, the $O \cap D$ is dense in $O$: Let $U$ be any non-empty open subset of $U$. Then $U$ is also open in $X$ (it is of the form open in $X$ intersected with $O$ so open), hence intersects $D$, so $\emptyset \neq U \cap D = (U \cap O) \cap D = U \cap (O \cap D)$, showing that $O \cap D$ is dense in $O$.

  3. Also if (for $A \subset B \subset X$) $A$ is dense in $B$ and $B$ is dense in $X$, $A$ is dense in $X$. Let $O$ be a non-empty open in $X$, $O \cap A$ is open and non-empty (as $A$ is dense) in $A$, so $\emptyset \neq (O \cap A) \cap B = O \cap (A \cap B)$, showing that $A \cap B =A$ is dense in $X$.

Now for your case: $A$ and $B$ both open and dense. The first fact gives $A \cap B$ is open, the second that $A \cap B$ is dense in $B$ and as $B$ is dense in $X$ ,$A \cap B$ is dense in $X$ by the third fact, as claimed.

Throughout I use that $D$ is dense in $X$ iff every non-empty open subset of $X$ intersects $D$ (non-emptily).

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A set is dense iff it intersects all non-empty open sets. So let $U$ be non-empty and open. We want to show it intersects $A\cap B$. As $A$ is dense, $A\cap U$ is non-empty. And as $A$ is open, $A\cap U$ is open. Hence $A\cap U$ is a non-empty open set, hence it intersects the dense set $B$. So $(A\cap B)\cap U=(A\cap U)\cap B$ is non-empty, i.e., $A\cap B$ is dense. We didn't use that $B$ is open so far. We need it only to see that $A\cap B$ is again open. In other words,

Let $A,B$ be dense subsets of $X$. If at least one of $A,B$ is open, then $A\cap B$ is dense.