Question

Suppose that A⊂B⊂X and A is dense in closure of B and B is dense in X.
Then A is also dense in X.

Answer 1

Answer on Question #42210 – Math – Topology

Question. Suppose that $A \subset B \subset X$ and $A$ is dense in closure of $B$ and $B$ is dense in $X$. Then $A$ is also dense in $X$.

Proof. Recall that a subset $A \subset X$ is dense if for every open $U \in X$ the intersection $A \cap U \neq \emptyset$.

Also recall that a closure, $\overline{A}$, of $A$ in $X$ is the set of all $x \in X$ such that for every open $V \subset X$ containing $x$ the intersection $A \cap V \neq \emptyset$.

First we establish the following lemma:

Lemma. A is dense in $X$ if and only if $\overline{A} = X$.

Proof. Suppose $A$ is dense in $X$. Let $x \in X$ and $V \subset X$ be any open subset containing $x$. Then $A \cap V \neq \emptyset$ as $A$ is dense in $X$, and so $x \in \overline{A}$, that is $\overline{A} = X$.

Conversely, suppose $\overline{A} = X$. Let $U \subset X$ be any open non-empty set and $x \in U$. As $x \in \overline{A} = X$, we have that $U \cap A \neq \emptyset$, and so $A$ is dense in $X$.

Now the proof is easy. Due to lemma the assumption that $A$ is dense in closure of $B$, means that $\overline{A} = \overline{B}$, while the assumption that $B$ is dense in $X$ means that $\overline{B} = X$. Thus

$$\overline{A} = \overline{B} = X,$$

and so $A$ is dense in $X$.

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