# Definition of a Topology from Metric Spaces

In Rudin's Real and Complex Analysis, he give a motivation for the definition of a topology using open sets in a metric space. He says, for a metric space $$X$$, and for a set $$\tau$$ of sets $$E \subset X$$ if

$$(A)$$ $$\tau$$ is the set of opens sets of $$X$$ (defined using neighborhoods and interior points)

then

$$(B)$$ $$\tau$$ is a topology. i.e.

$$\,\,\,\,\,\,\,\,(i)$$ $$\,\,\,\,\emptyset \in \tau$$, $$X \in \tau$$

$$\,\,\,\,\,\,\,\,(ii)$$ $$\,\,\,\,\tau$$ is closed under countable intersections

$$\,\,\,\,\,\,\,\,(iii)$$ $$\,\,\,\,\tau$$ is closed under countable and uncountable unions.

I understand the direction $$(A)\implies(B)$$. But is it necessarily true that if $$\tau$$ is a topology on a metric space $$X$$, then $$\tau$$ is the set of open sets. i.e. $$(B)\implies(A)$$. Or similarly, does $$(B)\implies (A')$$ where

$$(A')$$ $$\tau$$ is a set of open sets of $$X$$. (not the set of open sets).

If $$(B)$$ does not imply $$(A)$$ or $$(A')$$ then why do we use $$(B)$$ as the topological definition for open sets if it permits sets other than open in a metric space?

• No, you can have different topologies on a metric space (for example, every set of any type, i.e. any set of points, can have the discrete topology on it). By different, I mean a topology that doesn't have the same open sets as the metric topology Mar 27, 2020 at 21:46
• @rubikscube09 so you can have a topology where the "open sets" are closed sets of $X$?
– BENG
Mar 27, 2020 at 21:48
• a topology has to satisfy certain axioms; e.g., any union of members of the topology still belongs to it; that is not satisfied for sets that are closed in the usual topology on $\mathbb R$; consider $\large{\cup}_{n=1}^\infty [\frac1n, 1-\frac1n]=(0,1)$ Mar 27, 2020 at 21:51
• "so you can have a topology where the "open sets" are closed sets of X" No, because the closed sets are not closed under infinite unions. Mar 27, 2020 at 21:53
• A topology is, by def'n, closed under finite intersections and arbitrary unions, but not necessarily under countable intersections. Mar 28, 2020 at 8:46

There are many ways of defining a topology on a set and, in particular, on a metric space. For instance, in $$\mathbb R$$, if $$\tau$$ consists of $$\emptyset$$, $$\mathbb R$$ and every interval of the form $$(-\infty,a)$$ or $$(-\infty,a]$$, then $$\tau$$ is another topology. Note that (in this example) some elements of $$\tau$$ are not open sets with respect to the usual topology on $$\mathbb R$$. Also, some open subsets with respect to the usual topology on $$\mathbb R$$ do not belong to $$\tau$$.

No. $$B\not \implies A$$.

The Euclidean metric is ONE possible topology on $$R$$ but it is not the only one. Any set of sets to satisfy B) will be a topology.

Consider the discrete topology where every set is both open and closed. $$\tau =$$ then set of all subsets. B) is certainly satisfied. $$\emptyset \in \tau$$ and $$\mathbb R \in \tau$$ and the intersection and union of any countable or uncountable union are intersection is a set.

Or consider the topology where $$\tau = \{\emptyset, \mathbb R\}$$ and $$\emptyset$$ and $$\mathbb R$$ are the only sets that are open or closed. B) is satisfied as $$\emptyset\cap \emptyset = \emptyset, \emptyset \cup \emptyset = \emptyset, \emptyset \cap \mathbb R = \emptyset, \emptyset \cup \mathbb R = \mathbb R, \mathbb R \cap \mathbb R = \mathbb R; \mathbb R \cup \mathbb R = \mathbb R$$.

Another not so trivial one is $$\tau = \{\emptyset, \mathbb R\}\cup \{(-\infty, x)|x\in \mathbb R\}$$.

=

• However the discrete topology can be derived from the discrete metric. Mar 27, 2020 at 22:09
• Good point. That's true. I was having a brain fart and thinking the OP was only talking of the Euclidean metric and equivalent. Mar 27, 2020 at 22:17