Metric space: Convergence and continuity of metric spaces


In metric space, we can not talk about convergence and continuity without applauding Felix Hausdorff for his immense contributions in set theory, measure spaces, topology and metric spaces. His contributions in metric spaces lead to the discovery of continuity and convergency of a sequence in a metric space. 
To talk about convergence then we must be well equipped with the knowledge of sequence of real numbers because it is the metric on $\mathbb{R}$ that enables us to define the basic concept of convergence of such sequence. In the case of sequence of complex numbers, we have to define the metric on the complex plane, while in the arbitrary metric space $X=(X,d)$ we consider a sequence $(x_n)$ of elements $x_1,x_2,...$ of $X$ and use the metric $d$ to define the convergence. For the definitions of convergence and continuity, we will take a look at one properties of each definition which will be followed by proof. Remember, theorems and proposition gives a better understanding of mathematical concepts. In our next class we will take a look at more theorems and propositions. 

Convergence of sequence in a metric space
A sequence in a metric space $X=(X,d)$ is said to converge if there is an $x\in{X}$ such that:
$\lim_{n\rightarrow\infty}d(x_n,x)=0$, $x$ is a limit of $x_n$ and we write $\lim_{n\rightarrow\infty}x_n=x$ for every $\epsilon>0$, $\exists{N},n\geq{N}\Rightarrow{x_n}\in{B}_\epsilon(x)$. 
We say $x_n$ converges to $x$ or has the limit $x$. If $x_n$ does not converges then we say it diverges.
Now let's take a look at a proposition and proof to gain more insight on convergence of metric space. 

Proposition
in a metric space, a sequence $(x_n)$ can only converge to one limit, denoted by $\lim_{n\rightarrow\infty}x_n$. 
Proof: Suppose $x_n\rightarrow{x}$ and $x_n\rightarrow{y}$ as $n\rightarrow\infty$ with $x\neq{y}$. Then they can be seperated by two disjoint balls $B_r(x)$ and $B_r(y)$ and a criteria of convergence is $\exists{N_1}, n\geq{N_1}\Rightarrow{x_n}\in{B_r}(x)$, 
$\exists{N_2}, n\geq{N_2}\Rightarrow{x_n}\in{B_r}(y)$. For $n\geq\max{N_1,N_2}$ which will result in $x_n\in{B_r(x)\bigcap{B_r(y)}}=\phi$ which is a contradiction.  $\square$. 

Continutiy of metric spaces. 
A function $f:X\rightarrow{Y}$ between metric spaces is conitnuous when it preserves convergence, i.e $x_n\rightarrow{x}$ in $X$ $\Rightarrow$ $f(x_n)\rightarrow{f(x)}$ in $Y$. 

Proposition 
If $f:X\rightarrow{Y}$ and $g:Y\rightarrow{Z}$ are continuous, so is $g\circ{f}:X\rightarrow{Z}$. 
Proof: to proof this, let's use continuity, let $x_n\rightarrow{x}$ in $X$, by continuity of $f$, $f(x_n)\rightarrow{f(x)}$ in $Y$. By coninuity of $g$, $g\circ{f(x_n)}=g(f(x_n))=g\circ{f(x)}$ in $Z$. 
Alternatively, let $W$ be any open set in $Z$. Then $g^{−1}W$ is an open set in $Y$, and so
$f^{−1}g^{−1}W$ is an open set in $X$. But this set is precisely $(g\circ{f})^{−1}W$ $\square$.