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

#### 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 some fuuther 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:

" ... ...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

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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

#### Oxide

##### New member
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
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