I am new to phrasing things as is done in topology so was hoping someone can tell me if my interpretation is correct here:
To prove a subset $A$ of a topological space $X$ is dense in $X$ one must show that for all $x$ in $X$ such that $x \not\in A$ there exists a function $f(x,N) \in A$ such that $\lim _{N\rightarrow \left| A \right| }(f \left( x,N \right)) =x$
For example the rational numbers are a dense subset of the real numbers because every real number either is a rational number or has a rational number arbitrarily close to it, for this case $f$ would be Diophantine approximation, and demonstrating this to be capable of approximating any irrational number to any desired precision with a rational algebraic expression is hence effectively proving that $\mathbb Q$ is a dense subset of $\mathbb R$.
Thanks for helping in advance
Best Answer
To show a subset $A$ of $X$ is dense (in general), we need to show that every non-empty (basic) open subset $O$ of $X$ contains a point of $A$. So for the reals it suffices to note that in any open interval $(a,b)$ with $a < b$ contains a rational number. We don't have to find some proscribed approximating function etc. And a set can be dense but have no sequences in it converge to points outside it, even in pseudocompact normal spaces. Dense is just that every non-empty open set intersects it, so in that sense the set "lives everywhere". In a metric space we can show that $A$ is dense iff for all $x \notin A$ there is a sequence $(a_n)_n$ from $A$ converging to $x$, and so maybe that's where the OP gets his inspiration.