1) If
is open then every point of
is an interior point of
. Clearly then no point of
may be a boundary point of
so that it follows that
. Conversely, suppose that that
, then no point of
is a boundary point of
. Thus, for every
we must have that there exists some neighborhood of
, which we'll call
such that either
or
. But, since
the latter is impossible and the conclusion follows.
2) If a set
is closed then it contains all it's limit points. So let
. If
is in
we are done so assume that
. Since
we must have that every neighborhood
contains points of both
and
, and since
we must conclude that the points of
are distinct from
. But, this implies that
is a limit point of
and the conclusion follows. Conversely, suppose that
. Let
be a limit point of
. If
we are done, so assume that
. Then every neighborhood of
contains points of
but since every neighborhood of
also contains
it follows that
. We may therefore conclude that
contains all it's limit points. The conclusion follows.