Topology induced by a metric

Stub grade: A*

This page is a stub

This page is a stub, so it contains little or minimal information and is on a to-do list for being expanded.The message provided is:

Demote to grade B once there are links in place to this page

Definition

Let $(X,d)$ be a metric space. Then there is a topology we can imbue on $X$, called the metric topology that can be defined in terms of the metric, $d:X\times X\rightarrow\mathbb{R}_{\ge 0}$.

We do this using the concept of topology generated by a basis. We claim ("Claim 1"):

• $\mathcal{B}:\eq\left\{B_r(x)\ \vert\ x\in X\wedge r\in\mathbb{R}_{>0} \right\}$ satisfies the conditions to generate a topology (and is a basis for that topology) - where $B_\epsilon(p)$ denotes the open ball of radius $\epsilon\in\mathbb{R}_{>0}$ centred at $p\in X$

The resulting topological space, say $(X,\mathcal{ J })$, has basis $\mathcal{B}$

• Explicitly the topology is $$\mathcal{J}:\eq\left\{\left. \bigcup_{B\in\mathcal{F} }B\ \right\vert\ \mathcal{F}\in\mathcal{P}(\mathcal{B})\right\}$$
  • Notice $\bigcup_{B\in\emptyset} B\eq\emptyset$ - hence the empty-set is open - as required.
  • Notice also that $\bigcup{B\in\mathcal{B} }B\eq X$ - obvious as $\mathcal{B}$ contains (among others) an open ball centred at each point in $X$ and each point is in that open ball at least.
    TODO: Copy and paste a proof from elsewhere

Consequences

• The open sets are exactly those which are unions of basis elements. i.e.

Grade: A*

This page requires some work to be carried out

Some aspect of this page is incomplete and work is required to finish it
The message provided is:

These should have more far-reaching consequences on the site. Like on the neighbourhood page and such

Proof of claims

• Claim 1: $\mathcal{B}$ satisfies the conditions to generate a topology for which it is a basis - see The set of all open balls of a metric space are able to generate a topology and are a basis for that topology