Let E be a subset of R n
. show that the derived set of E, E’, is closed
Solution
Suppose that X is a space such that all singleton subsets of X are closed.
(this includes all metric spaces; this condition is known
by the name T_1).
Then for all subset E of X, E is closed, where A' is the derived set of E.
Recall that E' is the set of all x such that every open neighbourhood of x
(every open ball, if you are working metrically) intersects E in a point other than x.
Proof: Suppose that x is NOT in E'. This means that there is an open neighbourhood
U of x, such that U / E is either {x} or empty.
Now, every point y of U is also NOT in E': for x this is clear, for all
other y we use U{x}, which is open as X has closed singletons
(open minus closed is open) and is a neighbourhood of y that misses E,
and so shows that y is not in E'.
So U misses E', and this shows that for every point of XE' we can find
a neighbourhood of it that is also inside XE'.
So XE' is open and E' is closed.