# Convergence in a Topological Space

In a topological space \( X \), a point \( x \in X \) is called a **limit point** of the sequence \( (x_n) \) if, for every neighborhood \( U \) of \( x \), there exists a positive integer \( N \) such that for all \( n \geq N \), \( x_n \in U \).

In other words, the sequence \( (x_n) \) converges to \( x \) if, from some index \( N \) onward, all terms of the sequence lie within every neighborhood of \( x \).

Mathematically, this is expressed as:

$$ \lim_{n \to \infty} x_n = x $$

In this case, \( x \) is called the **limit point** of the sequence \( (x_n) \).

## A Practical Example

Let's look at a concrete example with the sequence \( \frac{1}{x_n} \) in the topological space \( X = \mathbb{R} \) with the standard topology.

$$ x_n = \left( \frac{1}{n} \right) $$

I want to show that the sequence \( \left( \frac{1}{n} \right) \) converges to 0. That is, I want to prove that 0 is the limit point of the sequence \( \left( \frac{1}{n} \right) \).

To do this, I take any neighborhood \( U \) of 0.

In the standard topology of \( \mathbb{R} \), a neighborhood \( U \) of 0 contains an open interval of the form \( (-\epsilon, \epsilon) \) for some \( \epsilon > 0 \).

I need to find a positive integer \( N \) such that for every \( n \geq N \), \( \frac{1}{n} \in (-\epsilon, \epsilon) \).

Given \(\epsilon > 0\), I can choose \( N = \left\lceil \frac{1}{\epsilon} \right\rceil \). Therefore, for every \( n \geq N \), we have:

$$ n \geq \frac{1}{\epsilon} \implies \frac{1}{n} \leq \epsilon. $$

Consequently, \( \left| \frac{1}{n} \right| < \epsilon \) for every \( n \geq N \).

Therefore, for every neighborhood \( U \) of 0, there exists an \( N \) such that for every \( n \geq N \), \( \frac{1}{n} \in U \). This proves that \( \lim_{n \to \infty} \frac{1}{n} = 0 \).

In summary, \( 0 \) is the limit point of the sequence \( \left( \frac{1}{n} \right) \).

$$ \lim_{n \rightarrow \infty} \frac{1}{n} = 0 $$

Simply put, the sequence \( \frac{1}{n} \) converges to zero because, from a certain point onwards, all terms of the sequence fall within any neighborhood of zero.

Here is a table with the first ten values of the sequence:

$$

\begin{array}{|c|c|}

\hline

n & \frac{1}{n} \\

\hline

1 & 1 \\

2 & 0.5 \\

3 & 0.333 \\

4 & 0.25 \\

5 & 0.2 \\

6 & 0.167 \\

7 & 0.143 \\

8 & 0.125 \\

9 & 0.111 \\

10 & 0.1 \\

\hline

\end{array}

$$

For example, if I take \( N=5 \) where \( x_5= \frac{1}{5} = 0.2 \), for every \( n>5 \), the subsequent points belong to the same neighborhood \( U=(0 , 0.2) \).

The same happens if I take any other value of \( N \).

For instance, if I consider \( N=10 \) where \( x_{10}=0.1 \), for every \( n>10 \), the sequence is contained within the neighborhood \( U = (0, 0.1) \), and so on.

Therefore, zero is the limit point of this sequence.

And so on.