Convergence of a Sequence - Topological Version

Note

This is a version of convergence using the Epsilon-Neighborhood of some $a \in R$ . If you want a bit more background see Convergence of a Sequence.;

Convergence of a Sequence: Topological Version

A sequence $(a_{n})$ converges to $a$ if, given any $ε$ -neighborhood $V_{ε} (a)$ for $a$ there exists a point in the sequence after which all terms are in $V_{ε} (a)$ .

In other words, every $ε$ -neighborhood contains all but a finite number of terms of $(a_{n})$ .

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This definition is saying the same thing as Convergence of a Sequence#^09d171; $N$ in the original version dictates the point $a_{N}$ such that any $a_{n}$ past $a_{N}$ is in $V_{ε} (a)$ . Notice again that the value of $N$ is dependent on $ε$ . The smaller $ε$ is, the larger $N$ tends to have to be.

Example

Consider $(a_{n}) = \frac{1}{\sqrt{n}}$ . Show that $a_{n} \to 0$ .

Doing some scratch, we want to first have $ε > 0$ . Then $V_{ε} (a) = V_{ε} (0) = (- ε, ε)$ . We want to show that $\exists N$ where for all $n \geq N$ that $a_{n} \in V_{ε} (0)$ .

To show an example, first say that $ε = 1 / 10$ . What $N$ do we need to use? We need, at this point:

a_{n} \in (\frac{- 1}{10}, \frac{1}{10})

after $n \geq N$ . We need:

\frac{1}{\sqrt{N}} = a_{N} \in (\frac{- 1}{10}, \frac{1}{10}) \Rightarrow \frac{- 1}{10} < \frac{1}{\sqrt{N}} < \frac{1}{10}

Thus then since $\frac{- 1}{10} < 0 < \frac{1}{\sqrt{N}}$ for any $N$ then we ignore the left inequality, so:

\frac{1}{100} > \frac{1}{N} \Rightarrow N > 100

so choose $N = 101$ and then $n \geq N$ will also satisfy this result. In general, if we use $ε$ instead of $\frac{1}{10}$ then:

ε^{2} > \frac{1}{n} \Rightarrow n > \frac{1}{ε^{2}}

So use $N > \frac{1}{ε^{2}}$ . Then $n \geq N > \frac{1}{ε^{2}}$ will give the reverse results that $a_{n} \in V_{ε} (0)$ .

Proof
Let $ε > 0$ . Choose $N \in N$ such that:

N > \frac{1}{ε^{2}}

Then $n \geq N > \frac{1}{ε^{2}}$ so:

\begin{aligned} n & > \frac{1}{ε^{2}} \\ ε & > \frac{1}{\sqrt{n}} \\ ε & > | \frac{1}{\sqrt{n}} - 0 | \end{aligned}

Thus $a_{n} \in V_{ε} (0)$ .
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