Why is dense metric space S of M not equal to M?

Question

If $S$ is dense in $M$. For any $x \in M$; B$(x,r)\cap S$ is non empty for any $r$. Then why is $S$ not equal to $M$?

Answer 1

We can see an example with $\mathbb{Q}$ and $\mathbb{R}$. We have $\mathbb{Q}$ dense in $\mathbb{R}$, but $\mathbb{Q} \neq \mathbb{R}$. Why is it like this? Because $\mathbb{R}$ has numbers that $\mathbb{Q}$ doesn't have. Take for example $\pi$, which is an irrational number. $\pi = 3,141592...$

$\mathbb{Q}$ is dense in $\mathbb{R}$, hence for any $r$ we have that $B_r(\pi) \cap \mathbb{Q} \neq \emptyset$, where $B_r(\pi)$ is the ball in $\mathbb{R}$ with radius $r$ centered in $\pi$. For example, if we take $r = \frac{1}{100}$, we have that $B_r (\pi)$ is the ball of every number with distance less than $\frac{1}{100}$ from $\pi$. Hence $\frac{3141}{1000} = 3,141 \in B_{\frac{1}{100}}(\pi)$.

$\pi$ is a practical example of a point which is in $\mathbb{R}$ but not in $\mathbb{Q}$. $\pi$ is surrounded by points in $\mathbb{Q}$, but it is not itself in $\mathbb{Q}$.

However, you can also say that $\mathbb{R}$ is dense in $\mathbb{R}$, and in this special case the spaces are the same.

Answer 2

Example: $M=\mathbb R, S= \mathbb Q$ and $d(x,y)=|x-y|.$

Then $S$ is dense in $M$, but $S$ is a proper subset of $M$.