Vtyjkyuk

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 $(star) hspace1mm 0 leqslant fracd(l,m)2 < epsilon Longrightarrow hspace1mm (star^prime) hspace1mm fracd(l,m)2 =0 Longrightarrow l = m.$

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

It’s badly stated. The point is that

$$0lefracd(ell,m)2<epsilon$$

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

One has the following statement:

Let $X$ be a Hausdorff space and $(x_n)_ninmathbbNin X^mathbbN$, then $(x_n)_ninmathbbN$ has at most one limit in $X$.

Assume that $(x_n)_ninmathbbN$ is convergent to $x_1in X$ and assume there exists $x_2in Xsetminusx_1$ such that $(x_n)_ninmathbbN$ is convergent to $x_2$. Since $x_1neq x_2$ 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 $V_1cap V_2=varnothing$. Since $(x_n)_ninmathbbN$ is convergent to $x_1$ there exists $N_1inmathbbN$ such that for all $ngeqslant N_1,x_nin V_1$, likewise there exists $N_2inmathbbN$ such that for all $ngeqslant N_2,x_nin V_2$. Define $N:=max(N_1,N_2)$, one has $x_Nin V_1cap V_2$, 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 $xneq y$, then one has:
$$d:=d(x,y)>0.$$
$V_x:=Bleft(x,fracd2right)$ is an open neighborhood of $x$ and $V_y:=Bleft(y,fracd2right)$ is an open neighborhood of $y$, one has:$$V_xcap V_y=varnothing.$$
Indeed, otherwise there exists $zin V_1cap V_2$ and one has:
$$d(x,z)<fracd2textrm and d(y,z)<fracd2.$$
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.

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.