**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
.