1
$\begingroup$

I'm reading through a proof in my lecture notes:

Every convergent sequence of a metric space has a unique limit.

Suppose $\lim_{n \to \infty} x_n = l$ and $\lim_{n \to \infty} x_n = m.$ Then for every $\epsilon > 0,$ there exists $N$ such that for every $n \geqslant N,$ $$d(l,m) \leqslant d(l,x_n) + d(x_n,m) < \epsilon + \epsilon = 2 \epsilon. $$

So $(\star) \hspace{1mm} 0 \leqslant \frac{d(l,m)}{2} < \epsilon \Longrightarrow \hspace{1mm} (\star^\prime) \hspace{1mm} \frac{d(l,m)}{2} =0 \Longrightarrow l = m.$

I can follow the proof up until the jump from $(\star)$ to $(\star^\prime)$. Why does $\frac{d(l,m)}{2}=0$?

$\endgroup$
4
  • $\begingroup$ You've successfully shown that $d(l,m) \leq 2\varepsilon$ for all $\varepsilon$. There is only one nonnegative number that is smaller than every positive number. $\endgroup$
    – user98602
    Commented Nov 8, 2015 at 23:54
  • $\begingroup$ Choose $\epsilon = d(\ell, m)/2$, then you have $d(\ell, m) < d(\ell, m)$. There is the contradiction. $\endgroup$
    – IAmNoOne
    Commented Nov 8, 2015 at 23:54
  • $\begingroup$ We've concluded that $0\leq\frac{d(l,m)}2<\varepsilon$. Since $\varepsilon$ is an arbitrary small quantity, that means that $d(l,m)$ is smaller than any positive quantity I can think of. Hence, it must be 0 $\endgroup$
    – EA304GT
    Commented Nov 8, 2015 at 23:56
  • $\begingroup$ My extraordinary professor Claude Chabauty called the first axiom in definition of metric ($d(x,y)=0\iff x=y$) "Axiome de séparation". This is what determine the uniqueness of the limit. $\endgroup$
    – Piquito
    Commented Nov 9, 2015 at 0:07

3 Answers 3

5
$\begingroup$

It’s badly stated. The point is that

$$0\le\frac{d(\ell,m)}2<\epsilon$$

for every $\epsilon>0$, which is possible only if $\frac{d(\ell,m)}2=0$: otherwise it would be less than itself!

$\endgroup$
1
  • $\begingroup$ Do you have any advice on understanding This Question? I would greatly appreciate it. $\endgroup$
    – M.A.
    Commented Nov 11, 2015 at 16:02
4
$\begingroup$

One has the following statement:

Let $X$ be a Hausdorff space and $(x_n)_{n\in\mathbb{N}}\in X^\mathbb{N}$, then $(x_n)_{n\in\mathbb{N}}$ has at most one limit in $X$.

Assume that $(x_n)_{n\in\mathbb{N}}$ is convergent to $x_1\in X$ and assume there exists $x_2\in X\setminus\{x_1\}$ such that $(x_n)_{n\in\mathbb{N}}$ is convergent to $x_2$. Since $x_1\neq x_2$ and $X$ is Hausdorff, there exists $V_1$ an open neighborhood of $x_1$ in $X$ and $V_2$ an open neighborhood of $x_2$ in $X$ such that $V_1\cap V_2=\varnothing$. Since $(x_n)_{n\in\mathbb{N}}$ is convergent to $x_1$ there exists $N_1\in\mathbb{N}$ such that for all $n\geqslant N_1,x_n\in V_1$, likewise there exists $N_2\in\mathbb{N}$ such that for all $n\geqslant N_2,x_n\in V_2$. Define $N:=\max(N_1,N_2)$, one has $x_N\in V_1\cap V_2$, a contradiction.

Now, notice that, one has:

Let $(X,d)$ a metric space, then $X$ is Haussdorf.

Let $(x,y)\in X^2$ such that $x\neq y$, then one has: $$d:=d(x,y)>0.$$ $V_x:=B\left(x,\frac{d}{2}\right)$ is an open neighborhood of $x$ and $V_y:=B\left(y,\frac{d}{2}\right)$ is an open neighborhood of $y$, one has:$$V_x\cap V_y=\varnothing.$$ Indeed, otherwise there exists $z\in V_1\cap V_2$ and one has: $$d(x,z)<\frac{d}{2}\textrm{ and }d(y,z)<\frac{d}{2}.$$ Using triangle inequality it follows: $$d=d(x,y)<d,$$ which is a contradiction.

N.B. This answer is only here to show you that uniqueness of a limit of a sequence is not only true in metric spaces which are a particular case of Hausdorff spaces.

$\endgroup$
0
$\begingroup$

If $d(l,m)/2<\epsilon$ for every $\epsilon>0$, then $d(l,m)/2=0$ otherwise $d(l,m)/2<d(l,m)/2$ which is a contradiction.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .