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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..

Akash Patalwanshi
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  • I don't known, why negative votes? I am quite good in Real analysis. But question is for "general topological space" and I am beginner in topology. So one can have questions like this. Really beginner may face downvoting :-( – Akash Patalwanshi Mar 23 '18 at 05:20
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    I've just upvoted you :) some advice for the future: write it in a "mathematical" language and it will be much more approved :) – Riccardo Ceccon Mar 23 '18 at 10:20

5 Answers5

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  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.

Henno Brandsma
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  • Wait, does the bar over a set mean the closure or the complement of the set? If it's the complement, I don't believe that first statement is true. for $A = {1,2}, B = {1,2,3}, X = {1,2,3,4}$ then $\overline{A} = {3,4}, \overline{B} = {4}$ and so $\overline{A} \not\subset \overline{B}$ – Shufflepants Mar 23 '18 at 14:35
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    @Shufflepants closure not complement. – Henno Brandsma Mar 23 '18 at 14:45
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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.

Ross Millikan
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    You cannot always find such a sequence (in general spaces), but just use that $A \subseteq B$ implies $\overline{A} \subseteq \overline{B}$. If $A$ is dense, its closure is $X$ and thus so is the closure of any larger set. – Henno Brandsma Mar 23 '18 at 06:19
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Only if the family is non-empty, but if so, then it follows because any superset of a dense subset is dense.

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

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

Xin Fu
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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 $A'$.

Joonas Ilmavirta
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