Union of dense is dense?

Question

Question:In topological space, union of any family of dense subset is dense?

I don't know whether the above statement is true or not! I know the definition of dense sets in topological space. According to "me it may not be true, as closure of infinite Union may not be equals to Union of closures. please help me to prove the above or give counter example of above..

Answer 1

1. For all subsets $A,B$ of $X$: $A \subseteq B$ implies $\overline{A} \subseteq \overline{B}$.
2. $A$ is dense means $\overline{A} = X$.
3. So $A$ dense and $A \subseteq B$ then $B$ is dense.
4. $A_j \subseteq \bigcup_{i \in I} A_i$ for any $ j \in I$.

So if in a union of sets at least one of them is dense, so is the union.

Answer 2

Yes, it is true. Given one dense set you can find a sequence converging to any point of the space. Adding in more points to your set cannot remove any sequences, so you can still find a sequence converging to any point in the space. As an example, think of the rationals in $\Bbb R$. They are dense. Another dense set is the rationals times $\sqrt 2$. If I take the union of those, I can find a sequence converging to any real. I can just ignore the rationals times $\sqrt 2$ and use the sequence of rationals that I already had.

Answer 3

Suppose $\{A_n\}_{n=1}^\infty$ is a family of dense subsets in a topological space $X$, then for a fixed $n$, we have

$$A_n\subset\bigcup_{n=1}^\infty A_n\Rightarrow X=\overline{A_n}\subset\overline{\bigcup_{n=1}^\infty {A_n}}$$

Answer 4

Only if the family is non-empty, but if so, then it follows because any superset of a dense subset is dense.

Answer 5

It might be a good idea to look at alternative definitions of density. There are several, and some make this property more transparent than others. For example a set $A\subset X$ is dense iff for any non-empty open set $U\subset X$ we have $U\cap A\neq\emptyset$. That is, a set is dense when it meets every open set. Now if you take a bigger set $A'\supset A$ (whether it is a union or something else), then every open set has an element of $A$ and therefore an element of $A'$.