**Problem 1.** Let
be a topological
space; let
be a subset of
. Suppose that for
each
, there is an
open set
containing
such that
. Show that
is open in
.

*Solution.* For each
, choose an open
subset
such
that
. I claim that
.
Indeed, we have
for
each
by definition,
so
.
Furthermore, for each
, one has
,
which gives the opposite inclusion. As each
is open and
arbitrary unions of open sets are open, the union
is open in
and hence
is open in
.