Prove that there exists a sequence that contains convergent subsequences converging to any given real number.

Question

My attempt:

Let $x \in \mathbb{R}$. By the density of rationals there is an interval $(x, x+\frac{1}{n}$) that contains a rational number. Any value of $n$ will produce a real number and we are done.

Am I on the right track?

Answer 1

Here's an easier approach: Since $\mathbb N\cong\mathbb Q\times\mathbb N$, there exists some bijection $b:\mathbb N\to\mathbb Q\times\mathbb N$. Let $f:\mathbb Q\times\mathbb N\to\mathbb Q$ be $(a,b)\mapsto a$.

So, consider the sequence $x_n=f\circ b\;(n)$. Since every rational appears infinitely many times in $x_n$, you can find any sequence of rationals as a subsequence, and hence you are done (since every real is approached by some sequence of rationals).

Your approach doesn't work, since the sequence can't depend on a given real.