Topology of Plane Sets of Points 107The set UaEA Nv(a) is an open covering of A. Hence it contains a finite
subcovering, say
n
LJ N 11 ;(a;)
i=l
Let 8 = minv; (i = l, .... ,n). Then 8 > 0, and
n n
N5(b) n Ac N5(b) n LJ N 11 .(a;) = LJ[N5(b) n N 11 .(a;)] = 0.
i=lso that N5(b) CA'. Since bis any point of A', this proves that.A' is open,
so A is closed.
Next we show that A is bounded. Clearly, UaEA N 1 (a) is an open cover-
ing of A (a covering by open neighborhoods of radius 1). Hence it contains
a finite subcovering, for instance,
m
Let μ = max d( a 1 , a;) ( i = 2, 3, ... , m). For a E A there is some i such
that a E N 1 (a;). Therefore, for that i we have
d( a 1 , a) ~ d( a 1 , a;) + d( a;, a) < μ + 1
and it follows that AC Nμ+ 1 (a1). Hence A is bounded.
The converse of Theorem 2.14 is not true in general. However, for the
metric spaces ]Rn the converse is valid. This gives an intuitive and useful
chaxl:tcterization of compact sets in Euclidean spaces and, in particular, in
(<C, d). First we need the following proposition.
Theorem 2.15 A closed subset of a compact set. is compact.
Proof Let A C S be compact, and suppose that B C A is .clos:d., Then
B' is open. Let_ UaEI Ga be any open covering of B. It follows that
(LJaEI Ga) U B' is an open covering of A, so it has a finite subcovering
(since A is compact). This finite subcovering of A will cover B also (since
B C A). If B' is a part of the finite subcovering, we may omit B' (which
cannot cover any points of B), and leave B covered by the union of a finite
number of sets GDI. This shows that B is compact.
Theorem 2.16 A subset of !Rn is compact iff it is closed and bounded.
Proof By Theorem 2.14 all we need to show is that any closed and bounded
subset of Rn is compact. Consider for simplicity the case n = 2 (the proof
in the general case is similar). Let A be a closed and bounded subset of
]R^2 • Then A is contained in some disk in JR^2 , and also in some closed square