Proof of: Because $I$ is dense in $R$ there exists a sequence $\{x_n\} \subset I$ with $\{x_n\} \to r$.

Question

In a passage from some school slides I have written:

consider an arbitrary $r \in Q$. Because $I$ is dense in $R$ there exists a sequence $\{x_n\} \subset I$ with $\{x_n\} \to r$.

This seems sensible but I am unsure where to start a proper proof.

Answer 1

Isn't this by definition?

For any $r\in R$, it is a limit point of $I$. So by definition there exists a sequence in $I$ with limit $r$. And then $r\in Q$ implies $r\in R$.

Am I missing something?

Answer 2

Using the definition you provided in the comments:

Consider any $r \in \Bbb R$. By the density of the rationals, we may construct a sequence as follows:

• Select a rational $x_1$ such that $r - 1 < x_1 < r$
• Select a rational $x_2$ such that $r - 1/2 < x_2 < r$
• $\vdots$
• Select a rational $x_n$ such that $r - 1/n < x_n < r$ for each $n \in \Bbb N$

The resulting sequence $\{x_n\}_{n=1}^\infty$ converges to $r$.