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

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,886
Hobart, Tasmania
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 some fuuther help in order to fully understand Example 4.1.1 ...


Example 4.1.1 reads as follows:


Singh - Example  4.1.1 ... .png



In the above example from Singh we read the following:

" ... ...Then the complement of \(\displaystyle \{ x_n \ | \ x_n \neq x \text{ and } n = 1,2, ... \}\) is a nbd of \(\displaystyle x\). Accordingly, there exists an integer \(\displaystyle n_0\) such that \(\displaystyle x_n = x\) for all \(\displaystyle n \geq n_0\). ... ... "


My question is as follows:


Why, if the complement of \(\displaystyle \{ x_n \ | \ x_n \neq x\) and \(\displaystyle n = 1,2, ... \}\) is a nbd of \(\displaystyle x\) does there exist an integer \(\displaystyle n_0\) such that \(\displaystyle x_n = x\) for all \(\displaystyle n \geq n_0\). ... ... ?


Help will be much 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:


Singh - Defn 1.2.5 ... ... NBD ... .png



Singh - 1 - Start of Chapter 4 ... PART 1 .png
Singh - 2 - Start of Chapter 4 ... PART 2 .png



Hope that helps ...

Peter
 

Oxide

New member
Mar 19, 2020
2
India
The set \(\displaystyle U = \mathbb R \setminus \{ x_n \mid x_n \neq x \}\) has countable complement, namely \(\displaystyle \{ x_n \mid x_n \neq x\}\), and is therefore open, and also contains \(\displaystyle x\), which makes it a nbhd of \(\displaystyle x\).
From the definition of convergence, there exists a positive integer \(\displaystyle n_0\) such that for all \(\displaystyle n \ge n_0\), \(\displaystyle x_{n} \in U\). This means that \(\displaystyle x_n \neq x\) is false, and therefore \(\displaystyle x_n = x\).
 
Last edited:

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,886
Hobart, Tasmania
The set \(\displaystyle U = \mathbb R \setminus \{ x_n \mid x_n \neq x \}\) has countable complement, namely \(\displaystyle \{ x_n \mid x_n \neq x\}\), and is therefore open, and also contains \(\displaystyle x\), which makes it a nbhd of \(\displaystyle x\).
From the definition of convergence, there exists a positive integer \(\displaystyle n_0\) such that for all \(\displaystyle n \ge n_0\), \(\displaystyle x_{n} \in U\). This means that \(\displaystyle x_n \neq x\) is false, and therefore \(\displaystyle x_n = x\).



Thanks Oxide ...

I really appreciate your help ...

Peter