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.

- Are the statements in the propositions conditional or absolute? If conditional, what are the hypotheses and conclusions? If absolute, can you rephrase the statement as a conditional statement?
- Are the proofs of these propositions direct or indirect?
- Did the proofs leave some steps for the reader to complete? If so, state what steps the reader is expected to complete. [Optional: complete these steps.]
- Indicate any parts of the arguments that you felt needed greater detail or better connection to the proofs. [Optional: Supply these details or suggest a better connection.]
- Consider A= {1}. Prove: A is not an open set.
- Consider {x: 0 < x <=
1}= (0,1]. Prove: (0,1] is not an open
set.

- Optional: Generalize (unify) the results in problems 5 and 6. Prove your generalization.
- Suppose f be a function, f : R -> R with f (x) = 2x + 8. Prove: If O is an
open set of real numbers, then f
^{-1}(O) is an open set.