Metric Spaces

There is a close relationship between topologies and metric spaces. We will see that every metric space directly induces a topology based on its metric. (from a CS point of view, this means topologies are more general than metric spaces).

Definition of Metric Space

We say that $(\mathcal{X},d)$ is a metric space if $\mathcal{X}$ is a set and $d$ a function $\mathcal{X}\times \mathcal{X}\to \mathbb{R}$ such that:

1. Distance from it self is zero $d(a,b)=0⟺a=b$
2. Symmetric function: $d(a,b)=d(b,a),\forall a,b\in \mathcal{X}$
3. Triangle inequality is satisfied: $\forall a,b,c\in \mathcal{X}:d(a,b)+d(b,c)\ge d(a,c)$

Induced topology

If we define the sets $B(c,r)=\left\{p\in \mathcal{X}\mid d(p,c)<r\right\}$ to be open, this induces a topology $(\mathcal{X},\mathbb{B})$ where $\mathbb{B}:=\left\{B(c,r)\mid c\in \mathcal{X},r\in \mathbb{R}\right\}$. This should be easy to verify, but it is mostly uninteresting and quite intuitive (i'm not reasoning like a mathematician now).

Connectedness of $\left[0,1\right]$

We define the subspace topology of $\mathbb{R}$ on it's subset $\left[0,1\right]$ we can prove that this is connected. This is uses the intermediate value theorem see Limiti#Weierstrass e Valore intermedio, the concept is very similar.

Let's assume the space is disconnected, so there exists two non empty disjoint sets $A,B$ such that $A\cup B=\left[0,1\right]$. We want to build a function that says this gives a contradiction. Let's take two points, $a\in A$ and $b\in B$. And let's build a function in the following way:

$$f(x)=\{\begin{array}{l}0\text{ if }x\in A\\ 1\text{ if }x\in B\end{array}$$

We observe that this function is defined for every point in the domain, so it is well defined, and it's continuous because we have that every set of the domain has a pre-image either $A,B,A\cup B,\varnothing$, that are all continuous. But if we have these assumptions we have that for all inputs either $A=\varnothing$ or $B=\varnothing$ because if it has some values $0$ or $1$ then by the intermediate value theorem it should take all the values. This contradicts the hypothesis the set is disconnected.