Basic Theorems Regarding Dense Sets in a Metric Space

# Basic Theorems Regarding Dense Sets in a Metric Space

Recall from the Dense Sets in a Metric Space page that if $(M, d)$ is a metric space then a set $S \subseteq M$ is said to be dense in $M$ if for all $x \in M$ and for all $r > 0$ we have that:

(1)\begin{align} \quad B(x, r) \cap S \neq \emptyset \end{align}

In other words, $S$ is dense in $M$ if every open ball contains points from $S$.

We will now look at some nice theorems regarding dense sets in a metric space.

Theorem 1: Let $(M, d)$ be a metric space and let $S \subseteq M$. Then $S$ is dense in $M$ if and only if $\bar{S} = M$. |

*Sometimes dense sets in $M$ are defined such that $\bar{S} = M$.*

**Proof:**$\Rightarrow$ Suppose that $S$ is dense in $M$. Then for all $x \in M$ and for all $r > 0$ we have that:

\begin{align} \quad B(x, r) \cap S \neq \emptyset \end{align}

- So let $x \in M$. Then from the equation above we see that every $x \in M$ is an adherent point of $S$ and so since $\bar{S}$ is the set of adherent points of $S$ we have that $\bar{S} = M$.

- $\Rightarrow$ Suppose that $\bar{S} = M$. Then every point in $M$ is an adherent point of $S$, and so for every $x \in M$ and for all $r > 0$ we have that:

\begin{align} \quad B(x, r) \cap S \neq \emptyset \end{align}

- But this shows that $S$ is dense in $M$. $\blacksquare$

Lemma 1: Let $(M, d)$ be a metric space and let $S, T \subseteq M$. If $S$ is open then $S \cap \bar{T} \subseteq \overline{S \cap T}$. |

**Proof:**Let $x \in S \cap \bar{T}$. Then $x \in S$ and $x \in \bar{T}$. Since $x \in \bar{T}$ we have that $x$ is an adherent point of $T$ and so for all $r > 0$ we have that:

\begin{align} \quad B(x, r) \cap T \neq \emptyset \quad (*) \end{align}

- Now since $S$ is open we have that $S = \mathrm{int} (S)$. So since $x \in S$ we have that $x \in \mathrm{int}(S)$ and so $x$ is an interior point of $S$. More importantly, there exists an $r_0 > 0$ such that:

\begin{align} \quad x \in B(x, r_0) \subseteq S \end{align}

- So $B(x, r_0) \cap S \neq \emptyset$. In fact, for all $r > 0$ we then have that $B(x, r) \cap S \neq \emptyset$ $(**)$. From $(*)$ and $(**)$ we see that

\begin{align} \quad B(x, r) \cap (S \cap T) \neq \emptyset \end{align}

- So $x$ is an adherent point of $S \cap T$, i.e., $x \in \overline{S \cap T}$. Thus:

\begin{align} \quad S \cap \bar{T} \subseteq \overline{S \cap T} \quad \blacksquare \end{align}

Theorem 2: Let $(M, d)$ be a metric space and let $S, T \subseteq M$. If $S$ and $T$ are dense in $M$ and if either $S$ or $T$ is open then $S \cap T$ is dense in $M$. |

**Proof:**Without loss of generality assume that $S$ is open. Since $S$ and $T$ are dense in $M$ we have that $\bar{S} = M$ and $\bar{T} = M$. Now by Lemma 1:

\begin{align} \quad \overline{S \cap T} \supseteq S \cap \bar{T} = S \cap M = S \end{align}

- So $S$ is contained in the closure of $S \cap T$. However, $S$ is dense in $M$. So every point in $M$ is an adherent point of $S$, and consequentially, every point in $M$ is thus an adherent point of $S \cap T$ so $\overline{S \cap T} = M$ and thus $S \cap T$ is dense in $M$. $\blacksquare$