1.79
1. Let
𝑥
∈
int
𝑆
.
Thus
𝑆
is a neighborhood of
𝑥
.
Therefore,
𝑇
⊇
𝑆
is a
neighborhood of
𝑥
, so that
𝑥
is an interior point of
𝑇
.
2. Clearly, if
𝑥
∈
𝑆
, then
𝑥
∈
𝑇
⊆
𝑇
. Therefore, assume
𝑥
∈
𝑆
∖
𝑆
which implies
that
𝑥
is a boundary point of
𝑆
. Every neighborhood of
𝑥
contains other points
of
𝑆
⊆
𝑇
. Hence
𝑥
∈
𝑇
.
1.80
Assume that
𝑆
is open. Every
𝑥
∈
𝑆
has a neighborhood which is disjoint from
𝑆
𝑐
. Hence no
𝑥
∈
𝑆
is a closure point of
𝑆
𝑐
.
𝑆
𝑐
contains all its closure points and is
therefore closed.
Conversely, assume that
𝑆
is closed.
Let
𝑥
be a point its complement
𝑆
𝑐
.
Since
𝑆
is closed and
𝑥 /
∈
𝑆
,
𝑥
is not a boundary point of
𝑆
.
This implies that
𝑥
has a
neighborhood
𝑁
which is disjoint from
𝑆
, that is
𝑁
⊆
𝑆
𝑐
. Hence,
𝑥
is an interior point
of
𝑆
𝑐
. This implies that
𝑆
𝑐
contains only interior points, and hence is open.
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 Fall '10
 Dr.DuMond
 Macroeconomics, Closed set, General topology, Boundary, interior point

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