Metric Space, Induced topology

Question

I'm trying to show that the metric topology is indeed a topology.

To do so, I want to show the following three statements are true:

1. $\emptyset$ and X are in $\tau$
2. finite intersections of open sets are open
3. arbitrary (finite, countable, uncountable ...) unions of open sets are open

I'm having trouble with part 1, that is, with what I think I need to do in order to show that $X\in\tau$.

How can I show that $\tau$ contains an open ball $B_{\epsilon}$(x) for every $x\in X$?

Answer 1

You start with a metric $d$ on a set $X$, without a topology.

You then define a topology $\mathcal{T}$ on $X$ by $$O \in \mathcal{T} \text{ iff } \forall x \in O: \exists r>0: B(x,r) \subset O\text{.}$$

Of course, one has to check that this fulfils the axioms for a topology.

To see that $X$ (what you asked about) is in $\mathcal{T}$: for any $x \in X$, we can pick any $r >0$ that we like (so e.g. $1$ will do, but $r=100$ is also OK) and the $B(x,r)$ is always a subset of $X$ (by definition, as $B(x,r) = \{ y \in X: d(x,y) < r \} \subset X$) so the condition is satisfied for any $r$.

The empty set if a bit more subtle: the condition is true vacuously (any statement that starts with $\forall x \in \emptyset$ is true, as there is no $x$ in the empty set to falsify the rest of the statement), so $\emptyset \in \mathcal{T}$.

The definition I gave is equivalent to the condition that a set is open iff it is a union of some family of open balls, which is also a common way to formulate this. Or, if the notion of a base was introduced, we take the set of open balls of $d$ as the base for a new topology, and then we check that the set of $d$-open balls satisfies the axioms for a base for a topology instead.

Answer 2

The induced topology is usually defined as the topology that has the open balls as a base. For every $x\in X$, let $B_x$ be the open ball of radius $1$ and "center" $x$, that is, the set of all points $y$ such that $d(x,y)\lt 1$. Then $X$ is the union of the $B_x$, since $x\in B_x$.

Answer 3

Disclaimer: This post is meant informative. Where necessary things are left out for clarity.

---

Most varieties of proves rely on the approach given by Hausdorff as follows.

The collection of open balls with center at some chosen point is a filter base, that is: $$\forall B_r(x),B_{r'}(x)\exists B_{r_0}(x): B_{r0}(x)\subseteq B_{r}(x)\cap B_{r'}(x)$$ (That will be our neighborhood base later: $\mathcal{B}(x)$)

The upper set induced by the filter base is a filter then: $$\mathcal{N}(x):=\uparrow\mathcal{B}(x)$$ (That will be our neighborhood filter later: $\mathcal{N}(x)$)

What remains for this to actually deserve the attribute neighborhood is the correlation to its chosen point and the interrelation between points.

Let me first state this in general: $$\forall N\in\mathcal{N}(x):\qquad x\in N$$ $$\forall N\in\mathcal{N}(x)\exists U\in\mathcal{N}(x):\qquad N\in\mathcal{N}(u),u\in U$$ In the case of metric balls that is: $$\forall N: x\in B_r(x)\subseteq N$$ $$\forall N: B_{r(u)}(u)\subseteq B_r(x)\subseteq N$$ (The latter then can be chosen to be the open neighborhoods: $U$)

All together is called the neighborhood system: $$x\mapsto\mathcal{N}(x)$$

There is more to read in the neighborhood attribute and in the definition via neighborhoods.