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)


$(B)$ $\tau$ 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?

  • $\begingroup$ 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 $\endgroup$ Mar 27, 2020 at 21:46
  • $\begingroup$ @rubikscube09 so you can have a topology where the "open sets" are closed sets of $X$? $\endgroup$
    – BENG
    Mar 27, 2020 at 21:48
  • $\begingroup$ 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 $\large{\cup}_{n=1}^\infty [\frac1n, 1-\frac1n]=(0,1)$ $\endgroup$ Mar 27, 2020 at 21:51
  • $\begingroup$ "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. $\endgroup$
    – fleablood
    Mar 27, 2020 at 21:53
  • 1
    $\begingroup$ A topology is, by def'n, closed under finite intersections and arbitrary unions, but not necessarily under countable intersections. $\endgroup$ Mar 28, 2020 at 8:46

2 Answers 2


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 $\tau = \{\emptyset, \mathbb R\}\cup \{(-\infty, x)|x\in \mathbb R\}$.


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

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