Intuition for dense sets. (Real analysis)
A set $D$ is dense in $\Bbb R$ if every non-empty open interval of $\Bbb R$ contains a point of $D$; in symbols, $D$ is dense in $\Bbb R$ if and only if $D\cap(a,b)\ne\varnothing$ whenever $a,b\in\Bbb R$ and $a<b$. The set $\Bbb Q$ of rationals is dense in $\Bbb R$: if $a$ and $b$ are real numbers, and $a<b$, there is always a rational number between $a$ and $b$, so $\Bbb Q\cap(a,b)\ne\varnothing$. There is also always an irrational number between $a$ and $b$, so $\Bbb R\setminus\Bbb Q$, the set of irrationals, is also dense in $\Bbb R$. And of course $\Bbb R$ itself is dense in $\Bbb R$.
Another example of a dense subset of $\Bbb R$ is $\Bbb R\setminus\Bbb Z$, the set of real numbers that are not integers: you can easily prove that if $a<b$, the interval $(a,b)$ contains a non-integer. Similarly, $\Bbb R\setminus F$ is dense for any finite $F\subseteq\Bbb R$.
Here are a couple of less obvious examples. Let $$D=\left\{\frac{2m+1}{2^n}:n\in\Bbb N\text{ and }m\in\Bbb Z\right\}\;;$$ the elements of $D$ are the dyadic rationals, the rational numbers whose denominators in lowest terms are powers of $2$. This set $D$ is dense in $\Bbb R$; you might try to prove that $D\cap(a,b)\ne\varnothing$ whenever $a<b$. HINT: For a fixed $n$, the dyadic rationals with denominator $2^n$ are $\frac1{2^n}$ apart. Finally, let $C$ be the middle-thirds Cantor set; then $C$ does not contain any non-empty open interval, so $\Bbb R\setminus C$ intersects every non-empty open interval and is therefore dense in $\Bbb R$.
The prototypical example of a dense set is $\mathbb Q$ in $\mathbb R$. I can think of two ways of showing that a set $X$ is dense in $\mathbb R$:
- Show that it contains a subset $Y$ that is itself dense in $\mathbb R$.
- Show that every element of $\mathbb R$ is a limit of a sequence of elements in $X$.
EDIT: As suggested by Henry,
- Given any two distinct elements in $\mathbb R$, there is an element of $X$ between them.
Note that 1 applies to density in any topological space, 2 applies to any metric space, while 3 depends on the order of $\mathbb R$ is thus much less general.
A subset $A$ of ${\mathbb R}$ is dense in ${\mathbb R}$ if wherever you look in ${\mathbb R}$, you'll find an element of $A$ really close by. Formally, $A$ is dense means that every open interval $(a,b)$ of ${\mathbb R}$ contains an element of $A$.
How to prove that a given set $A$ is dense in ${\mathbb R}$ heavily depends on what $A$ is, but a normal strategy would be to start with an open interval $(a,b)$ of ${\mathbb R}$ and somehow explicitly construct an element of $A$ that's in there.