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