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Characterizing real intervals

Let .
• Let .
Then is either empty or or is the closed interval .
• Let .
Then is either empty or is or is the closed interval .
• Let .
Note that the set is empty (which is counted as an open set) if and only if has at most one element.
Specifically we have:
• , .
• If for some , then , and .
When has at least two elements, the sets , and are pairwise disjoint and have union .
Further , being the complement of the union of and is a non-empty open interval.
In particular, the set is always open, for any . Using the concept , we have the following theorem for a subset of the reals.
• The following are equivalent:
• If and with , then any with lies in the set .
Proof:
• If is the empty set, , then both conditions hold, since then and the second condition holds vacuously.
So henceforth, we may assume that is non-empty.
• First suppose that a non-empty set of reals has the property that whenever and , then any real with lies in .
We prove that .

So let .
Then, by definition, is neither an upper bound nor a lower bound for the set .
• Since is not an upper bound for , an element of must exist, with .
• Since is not a lower bound for , an element of exists, with .
Then we have and and , so .
So and we are done.
• Conversely, let be a non-empty set with .
Let and in be given, with .
Let obey .
We must prove that .

If or , there is nothing to do, since then .
So we may assume that .
• Then is not an upper bound for , since and .
• Also is not a lower bound for , since and .
Since is neither an upper bound nor a lower bound for the set , we have by definition of , so , so and we are done.
Now we use the theorem just proved to define an interval.
• A subset of the reals is said to be an interval if and only if either of the following equivalent conditions holds:
• .
• If and with , then any with lies in the set .
We want to determine all possible intervals.

We have the following classification theorem.
• Let be an interval (as just defined).
Then is one of the following ten types of sets:
• .
This is said to be open and closed and .
• .
This is said to be both open and closed and .
The open non-closed intervals:
• for some .
This is said to be open and not closed and .
• for some .
This is said to be open and not closed and .
• , for some and in , with .
This is said to be open and not closed and .
The closed non-open intervals:
• for some .
This is said to be closed and not open and .
• for some .
This is said to be closed and not open and .
• , for some and in , with .
This is said to be closed and not open.
• In the case that , the interval is and consists of a single point and then .
• Otherwise we have and then .
The intervals that are neither closed nor open:
• , for some and in , with .
This is said to be neither open nor closed and .
• , for some and in , with .
This is said to be neither open nor closed and .
We notice that if is any one of these ten kinds of sets, we always have and in all these ten cases, the set is the interior of : this is, by definition, the largest open subset of .

Proof of the classification theorem.
We need to prove:
• If is a subset of the reals, such that , then is one of the ten kinds of sets discussed above.
Proof:
• First, if , then is true and the set is in our list above, so we are done.
So henceforth we may assume that is non-empty.
• If has no upper or lower bound, then any real is neither an upper bound nor a lower bound for the set , so .
So we have .
• If is bounded below, but not bounded above, then exists.
Also , by definition of .
Then any real is neither an upper bound nor a lower bound for the set , so .
So we have or .
• If is bounded above, but not below, then exists.
Also , by definition of .
Then any real is neither an upper bound nor a lower bound for the set , so .
So or .
• If is bounded above and below, then exists, as does and we have .
Then , by the definitions of and .
• If now , we have , a closed interval.
So henceforth we may assume that , so .
• Then any real , such that is neither an upper nor a lower bound for the set , so we get that:
.
So we conclude that the set or or or .
In all cases, we have shown that is one of the kinds of sets in our list and the theorem is proved. We see also that the cases that turn up cover all ten possible kinds of non-empty intervals, as expected. Finally note the corollary:
• is an open interval if and only if .
If we do this, then we see that the complement of any interval is either an interval or a union of two intervals and the complement of an open interval is either a closed interval or the union of two closed intervals, whereas the complement of a closed interval is either an open interval or the union of two open intervals.

In general we define:
• An open subset of the reals is a subset that is a union of open intervals.
• A closed subset of the reals is a subset that is the complement of an open set.
The open and closed subsets form a topology for the reals:
• and are open.
• The union of any collection of open sets is open.
• The intersection of any finite collection of open sets is open.
Phrased in terms of closed sets the topology obeys:
• and are closed.
• The intersection of any collection of closed sets is closed.
• The union of any finite collection of closed sets is closed.

Next: Nested intervals Up: Thetheoryofnestedintervals Previous: Thetheoryofnestedintervals
George A. J. Sparling 2012-04-10