M. Flashman
More on Open Sets of Real Numbers
Reminder of the Definitions:
(1) For a and b real numbers with a < b, (a,b) ={ x : a < x
<
b}
(2) A set of real numbers, O, is called an open set if and only if
for any number x that is a member of O there are some numbers a
and b so
that x is a member of (a,b) and (a,b) is a subset of O.
Proposition 1: The empty set, Ø, is an open set.
Proof 1.1: Suppose Ø is not an open set. Then there is some number x that is a member of Ø and for any numbers a and b with x a member of (a,b), the set (a,b) is a subset of Ø. Thus if Ø is not an open set, Ø is not the empty set. Therefore ( by the contrapositive) the empty set is an open set. EOP.
Proof 1.2: Consider the open sets (0,1) and (3,4). We see
that
the intersection of (0,1) and (3,4) is the empty set. But the
intersection of two open sets is
an open set. Therefore the empty set is an open set. EOP.
Proposition 2: Suppose f be a function, f : R -> R with f (x) = 3x
+ 5. If O is an open set of real numbers, then f -1(O) is
an open set.
Proof: Suppose t
is an element of f
-1(O).
Let u = f (t) = 3t
+ 5 which is a member of O.
Since O is open , there are numbers c and d where u is a member of (c,d)
and (c,d) is a subset of O. Let a = (c-5)/3 and b = (d-5)/3. Then
t is a member of (a,b)
and (a,b) is a subset of f
-1(O), so f -1(O) is an open set. EOP.