Convergent sequence has one limit point

Question

I need to prove that if a sequence is convergent, it has exactly one limit point.

I have started proving this by contradiction, but can't seem to come to an equation that is obviously wrong. Please help.

What definition are you using for the limit point of the sequence? — John Habert, Feb 01 '14 at 18:07
d is a limit point if for $n \geq N$, $|a_n - d|\leq \epsilon$ — user989, Feb 01 '14 at 18:12
Suppose two limits $l_1\ne l_2$ and pick $\epsilon$ small enough. — Martín-Blas Pérez Pinilla, Feb 01 '14 at 18:15
I have written out the equation in this way, but don't see how this leads to a contradiction — user989, Feb 01 '14 at 18:16

score 4 · Answer 1 · answered Feb 01 '14 at 18:42

4

Suppose $\;|l_1-l_2|=\epsilon>0\;$ . Now:

$$\begin{cases}a_n\xrightarrow[n\to\infty]{}l_1\implies\;\exists\,N_1\in\Bbb N\;\;s.t.\;\;n>N_1\implies |a_n-l_1|<\frac\epsilon3\\{}\\a_n\xrightarrow[n\to\infty]{}l_2\implies\;\exists\,N_2\in\Bbb N\;\;s.t.\;\;n>N_2\implies |a_n-l_2|<\frac\epsilon3\end{cases}$$

So take $\;m>\max\{N_1,N_2\}\;$ , and then

$$\epsilon=|l_1-l_2|=|l_1-a_m+a_m-l_2|\le|l_1-a_m|+|l_2-a_m|<\frac23\epsilon...\text{contradiction.}$$

answered Feb 01 '14 at 18:42

DonAntonio

211,718
17
136
287

1

Why can we assume that the first line is valid? – user989 Feb 01 '14 at 21:25
2

It's a proof by contradiction, @user989 ... – DonAntonio Feb 02 '14 at 05:07

Convergent sequence has one limit point

1 Answers1