Convergence in Topological Spaces .... Singh, Example 4.1.1 .... ....

[Math Amateur]

TL;DR Summary

I need help in order to fully understand an example concerning convergence in the space of real numbers with the co-countable topology ...

I am reading Tej Bahadur Singh: Elements of Topology, CRC Press, 2013 ... ... and am currently focused on Chapter 4, Section 4.1: Sequences ...

I need help in order to fully understand Example 4.1.1 ...Example 4.1.1 reads as follows:

In the above example from Singh we read the following:

" ... ...no rational number is a limit of a sequence in $\mathbb{R} - \mathbb{Q}$ ... ... "My question is as follows:

Why exactly is it the case that no rational number a limit of a sequence in $\mathbb{R} - \mathbb{Q}$ ... ... "
Help will be appreciated ...

Peter=====================================================================================It may help readers of the above post to have access to Singh's definition of a neighborhood and to the start of Chapter 4 (which gives the relevant definitions) ... so I am providing the text as follows:

Hope that helps ...

Peter

 

[Infrared]

Suppose for contradiction that $\langle x_n\rangle$ is a sequence in $\mathbb{R}\setminus\mathbb{Q}$ that converges to a rational number $x$. Let $U=\mathbb{R}\setminus\bigcup_{n=1}^\infty \{x_n\}$. This is an open set in $\mathbb{R}_c$ because its complement is the set of the $x_n$, which is (at most) countable. Also $x\in U$ because $x$ is not any of the $x_n$ (since the $x_n$ are all irrational). So $U$ is an open neighborhood of $x$ that doesn't contain any of the $x_n$. This means that the $x_n$ do not converge to $x$.

 

[Math Amateur]

Suppose for contradiction that $\langle x_n\rangle$ is a sequence in $\mathbb{R}\setminus\mathbb{Q}$ that converges to a rational number $x$. Let $U=\mathbb{R}\setminus\bigcup_{n=1}^\infty \{x_n\}$. This is an open set in $\mathbb{R}_c$ because its complement is the set of the $x_n$, which is (at most) countable. Also $x\in U$ because $x$ is not any of the $x_n$ (since the $x_n$ are all irrational). So $U$ is an open neighborhood of $x$ that doesn't contain any of the $x_n$. This means that the $x_n$ do not converge to $x$.

Thanks for the help Infrared ...

Peter