How to connect the definitions of open sets and continuous functions w.r.t. metric space and topological space?

Question

For a topological space $(X,\tau)$, the topology $\tau$ on the set $X$ is a family of subsets called open sets, if $X$, $\emptyset$, any union of the subsets, and any finite intersection of the subsets are in $\tau$. This could be a definition of the term "open sets" with respect to topology.

For a metric space $(X,d)$, however, we have another definition of open sets, say, a subset $S \subset X$ is an open set if $\forall x \in S$, $\exists \epsilon > 0$ s.t. $B_{\epsilon}(x):=\{y \in X \;|\; d(x,y) < \epsilon\} \subset S$. This should be identical to the first definition in terms of metric topology. But how to show the second definition as a reduced version of the first one?

In addition, the concept of open sets seems strongly related to the concept of continuous functions. A generalized definition of the continuity in topological space is the inverse image of every open set is open. Traditionally, on the other hand, one defines the uniform continuity of a function $f(x)$ by $\forall x_1, x_2 \in X$, $\exists \delta > 0$ s.t. $\|f(x_1) - f(x_2)\| < \epsilon$ for $\forall \epsilon$ if $\|x_1 - x_2\| < \delta$. How to find the connections between the definitions in analysis and in topology?

Answer 1

For your first question, just take the collection of open sets in a metric space and show that they form a topology.

For your second question, first of all you gave the definition of continuity, not uniform continuity (which is stronger). The notion of continuity given in analysis in terms of epsilons and deltas is equivalent to the topological one in the case of metric spaces. Uniform continuity is, as far as I know, a notion proper of analysis, and not a topological one.

Answer 2

You may be overloading the word "open" here. You can talk about a topology without using the word "open" at all. A family $T$ of subsets of a set $X$ is a topology if $\emptyset \in T$, $X \in T$, the union of any family of sets in $T$ also belongs to $T$, and the intersection of any finite family of sets in $T$ also belongs to $T$.

If $X$ is a metric space, you can define a set $E$ to be open if for every point $x \in E$ there exists $\epsilon > 0$ so that $B_\epsilon (x) \subset E$. Now it is a matter of showing that the family of open sets forms a topology on $X$.

Answer 3

First, it's a standard exercise to check that metric-open sets form a topology in the general topology sense. Do it. The only axiom requiring a little thought is the finite intersection axiom.

Second, uniform continuity does not have a direct general topology formulation, for that we need another structure, a so-called uniform space. A uniform structure on a set gives us a topology on that set (in much the same way as metrics do) and for functions between uniform spaces we can define a notion of uniform continuity, which is stronger than continuity (between these sets as having the induced topologies).

Topological continuity in the open sets sense (inverse images of open sets are open) is (in a metric space) equivalent to regular continuity of $f$ at all points:

$$\forall x \in X: \forall \varepsilon > 0: \exists \delta >0 : \forall x' \in X: d_X(x,x') < \delta \implies d_Y(f(x), f(x')) < \varepsilon$$

for a function $f: (X,d_X) \to (Y, d_Y)$.