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.