# Dense Set

Let $X \subset \mathbb{R}$. A subset $S \subset X$ is called **dense in $X$** if any real number can be arbitrarily well-approximated by elements of $S$. For example, the rational numbers $\mathbb{Q}$ are dense in $\mathbb{R}$, since every real number has rational numbers that are arbitrarily close to it.

Formally, $S \subset X$ is dense in $X$ if, for any $\epsilon > 0$ and $x\in X$, there is some $s\in S$ such that $|x - s| < \epsilon$.

An equivalent definition is that $S$ is dense in $X$ if, for any $x\in X$, there is a sequence $\{x_n\} \subset S$ such that $\lim_{n\to\infty} x_n = x.$

$X$ that is the big rectangle. There is subset $S$ that can be said to be dense in $X$. This is because for any point $x\in X$, in this case a random point $x$ in the larger set $X$, one could draw a circle around $x$ using a random $s\in S$ as the radius and some element of that circle will be in $S$.

The image to the right shows this visually. There is a set## Examples of Dense Sets

The canonical example of a dense subset of $\mathbb{R}$ is the set of rational numbers $\mathbb{Q}$:

The rational numbers $\mathbb{Q}$ are dense in $\mathbb{R}$.

Take $x\in \mathbb{R}$. We may write $x = n + r,$ where $n \in \mathbb{Z}$ and $0 \le r < 1$. Consider the decimal expansion $r = 0. r_1 r_2 r_3 \cdots.$ Setting $x_k = n + 0.r_1 r_2 \cdots r_k,$ we see that each $x_k$ is rational and $\lim_{k\to\infty} x_k = x.\ _\square$

In fact, for any irrational number $\alpha \in \mathbb{R}$, the set $S_{\alpha} = \{a+b\alpha \, | \, a, b \in \mathbb{Z}\}$ is dense in $\mathbb{R}$. This is harder to prove than the above example, and requires clever use of the pigeonhole principle.

For any irrational $\alpha \in \mathbb{R}$, the set $S_{\alpha}$ is dense in $\mathbb{R}$.

Assume $\alpha > 0$. The proof when $\alpha < 0$ is entirely analogous.

Let $\{x\}$ denote the fractional part of $x$. First, we claim that the set $R_{\alpha} = \big\{\{n\alpha\} \, | \, n \in \mathbb{Z}\big\}$ is dense in $[0,1]$.

Choose $\epsilon > 0$ and pick an integer $m$ such that $m > \frac{1}{\epsilon}$. Divide the unit interval $[0,1]$ into $m$ subintervals, each of length $\frac1m$. By the pigeonhole principle, some two of the numbers $\{\alpha\}, \{2\alpha\}, \cdots, \{m\alpha\}, \{(m+1)\alpha\}$ must be in the same subinterval, so there exist integers $1\le i < j \le m+1$ such that $\Big|\{i\alpha\} - \{j\alpha\}\Big| < \frac1m$.

But $\Big|\{j\alpha\} - \{i\alpha\}\Big| = \{(j-i)\alpha\}$, so $\{(j-i)\alpha\} < \frac1m$. For any $y \in [0,1]$, there is some $0\le k \le m-1$ such that $y \in \left[\frac km, \frac{k+1}{m}\right]$. We may then pick an integer $q$ such that $q\{(j-i)\alpha\} = \{q(j-i) \alpha\} \in \left [ \frac{k}{m}, \frac{k+1}{m} \right],$ so that $\Big|y - \{q(j-i)\alpha\}\Big| < \frac1m < \epsilon$. Thus, $R_{\alpha}$ is dense in $[0,1]$.

Now, for any $z \in \mathbb{R}$, write $z = a + r,$ where $a \in \mathbb{Z}$ and $0\le r <1$. Since $R_{\alpha}$ is dense in $[0,1]$, there is $b \in \mathbb{N}$ such that $\Big|r - \{b \alpha\}\Big| < \epsilon$. It follows that $\Big|a- \lfloor b \alpha \rfloor + b\alpha - z\Big| = \Big|\{b\alpha\} - r\Big| < \epsilon.$ Since $a- \lfloor b \alpha \rfloor + b\alpha \in S_{\alpha}$, we conclude $S_{\alpha}$ is dense in $\mathbb{R}$. $_\square$

## Dense Sets in General Metric Spaces

One may define dense sets of general metric spaces similarly to how dense subsets of $\mathbb{R}$ were defined.

Suppose $(M, d)$ is a metric space. A subset $S \subset M$ is called

dense in $M$if for every $\epsilon > 0$ and $x\in M$, there is some $s\in S$ such that $d(x, s) < \epsilon$.

For example, let $\mathcal{C}[a,b]$ denote the set of continuous functions $f: [a,b] \to \mathbb{R}$. One may give $\mathcal{C}[a,b]$ the structure of a metric space by defining $d(f,g) = \max_{x\in [a,b]} \big|f(x) - g(x)\big|.$ By the extreme value theorem, this maximum exists for any two $f, g \in \mathcal{C}[a,b]$, so the distance function is well-defined. One can easily check that $d$ satisfies the axioms of a metric space.

Let $\mathcal{P} \subset \mathcal{C}[a,b]$ be the subset consisting of polynomial functions. The Stone-Weierstrass theorem states that $\mathcal{P}$ is dense in $\mathcal{C}[a,b]$. Intuitively, this means any continuous function on a closed interval is well-approximated by polynomial functions!