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

Peter

Well-known member
MHB Site Helper
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 ...

In the above example from Singh we read the following:

" ... ...no rational number is a limit of a sequence in $$\displaystyle \mathbb{R} - \mathbb{Q}$$ ... ... "

My question is as follows:

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

Help will be appreciated ...

Peter

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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

Last edited:

GJA

Well-known member
MHB Math Scholar
Hi Peter ,

Let $S=\{x_{n}\}_{n=1}^{\infty}$ be a sequence in $\mathbb{R}\setminus\mathbb{Q}$; i.e., a sequence of irrational numbers. Then the complement of $S$, $\mathbb{R}\setminus S$, is an open set in $\mathbb{R}_{c}$ containing the rationals. According to Definition 4.1.2, it follows that no point from $\mathbb{Q}$ could be the limit of $S$.

Peter

Well-known member
MHB Site Helper
Hi Peter ,

Let $S=\{x_{n}\}_{n=1}^{\infty}$ be a sequence in $\mathbb{R}\setminus\mathbb{Q}$; i.e., a sequence of irrational numbers. Then the complement of $S$, $\mathbb{R}\setminus S$, is an open set in $\mathbb{R}_{c}$ containing the rationals. According to Definition 4.1.2, it follows that no point from $\mathbb{Q}$ could be the limit of $S$.

Hi GJA ... ..

Thanks so much for the help ...

Peter