The Real Numbers have the Least Upper Bound
Property
Here, we prove that any set S ⊂ R bounded from above has a least upper
bound, i.e. sup(S) exists. In order to follow what is going on, you should
read this with a pencil in hand, and a sheet of paper. To get an idea of what
S is, just draw an open interval in R (note, this is not an accurate picture
of the situation, but it is a guide). Then, you should indicate where N is
(from the proof), and think about what our sequence Nk can look like. Think
about taking steps towards the set S from the number N .
At this level, it is not expected that you can create this proof yourself.
However, it IS expected that you can read it, follow it, understand it, and
hopefully pick up some tricks along the way. ENJOY!
Lemma 1: If S ⊂ R is bounded from above, and U is an upper bound such
1
that U 6= sup(S), then there is a smallest n ∈ N such that U − is an upper
n
bound.
1
First, we must show that there exists an n ∈ N for which U − is an
n
upper bound. Fortunately, since U is not the least upper bound of S, there
exists a number L such that L < U and L is an upper bound of S (otherwise,
U is the least upper bound!). Then, U − L = d is strictly positive. By the
Archemedian property of the Real numbers, there exists an n ∈ N such that
1
< d. But then,
n
1
U − >U −d=L
n
1
Hence, U − is an upper bound (because L is). Therefore, there exists an
n
1
n ∈ N such that U − is an upper bound of S. Now, the set {n ∈ N :
n
1
U − is an upper bound of S} is a nonempyt subset of the natural numbers.
n
Therefore, there is a smallest element. Hence, if S ⊂ R is bounded from
above, and U is an upper bound such that U 6= sup(S), then
min{n ∈ N : U −
1
is an upper bound ofS}
n
1
exists.
Now, we proceed with the proof.
Suppose S ⊂ R is bounded from above. If S is empty, then by convention,
we say sup(S) = −∞. Thus, we may assume S is nonempty and bounded
from above. Then, there exists an M ∈ R such that x ≤ M for all x ∈ S.
If M is the least upper bound, we are done. Otherwise, we continue our
search.
If M − 1 is not an upper bound of S, let N = M . Otherwise, let n0 be
the largest natural number such that M − n0 is an upper bound of S, and
define N = M − n0 .
Note that the above was merely cosmetic. We did all of this so N is an
upper bound of S such that N − 1 is not, i.e. there exists an x ∈ S such
that N − 1 < x. This is so that we can build a decreasing sequence of upper
bounds, and have some control over how fast the upper bounds converge to
their limit.
If N is the least upper bound, we’re done. Otherwise, let
1
n1 = min{n ∈ N : N − is an upper bound ofS}
n
1
and define N1 = N − . If N1 is our upper bound, we are done. Otherwise,
n1
we continue to look for a smaller upper bound. We make the following recursive definition to build a decreasing sequence of upper bounds with some
nice properties (more on this later).
If Nk−1 is the least upper bound, we’re done. Otherwise, let
1
nk = min{n ∈ N : Nk−1 − is an upper bound ofS}
n
1
and define Nk = Nk−1 − .
nk
Now, either Nk is the least upper bound for some k (and we’re done), or
we have a sequence of upper bounds {Nk }∞
k=1 .
2
Note that this is a very specific construction. It says: of all of the steps of
1
size you can take and still have an upper bound, take the biggest possible
n
step (a smaller n makes 1/n bigger, which is why we want the minimum of
the ns).
Now, there are two crucial aspects of this sequence. First, by construction, Nk+1 < Nk because we are subtracting a positive number from Nk to get
Nk+1 . This means that the sequence {Nk }∞
k=1 is strictly decreasing. Moreover, for any x ∈ S, Nk ≥ x for all k. Therefore, this sequence is bounded
from below (by this x in particular). By the completeness axiom of the real
numbers, any decreasing sequence bounded from below converges. Let
L = lim Nk .
k→∞
Now, we will come back to this in a minute. But first, we need the second
property of this sequence. This is what allows us to show that there is no
upper bound of S smaller than L:
Lemma 2: The sequence of nk is strictly increasing. Moreover,
k < nk < nk+1
(NOTE: these are the n corresponding to the step sizes. This says: each step
you take is shorter than the previous.)
Intuitively, the proof follows the idea: if the k + 1st step is bigger than
the k th step, our k th step could have been bigger. In terms of our setup, this
means that if nk+1 ≤ nk , then nk is NOT the smallest n from the set in the
definition of nk .
In order to keep this proof moving, I’ll prove this lemma at the end.
First, we have to show that in fact L is an upper bound. It is “obvious,”
but that isn’t a proof. The IDEA is that if L is not an upper bound, then
there is an x ∈ S with x > L. But then, there has to be some Nk between
L and x because Nk converges to L. This gives us that Nk is not an upper
bound of S, which is a contradiction. (The direct proof is more confusing).
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Suppose that L is NOT an upper bound. Then, there exists an x ∈ S
such that x > L. Let ² = x − L > 0. Since Nk converges to L, there exists
a K0 such that k ≥ K0 =⇒ |Nk − L| < ². However, Nk is decreasing, so
L < Nk and |Nk − L| = Nk − L. Therefore,
Nk − L < ² = x − L =⇒ Nk < x.
This is a contradiction to the fact that Nk is an upper bound of S. Thus,
there is no x ∈ S for which x > L, i.e. x ≤ L for all x ∈ S. This is precisely
the definition of an upper bound. Hence, L is in fact an upper bound.
Now, we must show that it is the LEAST upper bound. This was throwing me for a loop in class because I didn’t prove that the nk were increasing.
With this lemma, it isn’t hard at all. The problem was getting a handle on
what the nk are.
SIDETHOUGHT: It is not NECESSARY to prove that the nk are increasing. This is a stronger result than we actually need. It is only necessary
to prove that the nk approach infinity as k approaches infinity (do you know
what the difference is?). You should be able to convince yourself that they
must approach infinity (we don’t have the tools to use this as a proof). Here’s
why. N − 1 is NOT an upper bound. Therefore, N − 1 < L. Now, from the
constrution of the sequence Nk , we see that since
1
Nk = Nk−1 −
nk
we know
k
X
1
Nk = N −
n
j=0 j
Then, letting k go to infinity we have:
∞
X
1
N −1<L=N −
nk
k=0
so that
∞
X
1
<1
nk
k=0
From calculus, you were told that if a series converges, then the terms must
converge to 0 (you will be proving this later in the Advanced Calculus sequence). But, n1k → 0 if and only if nk → ∞ as k → ∞. So, the fact that the
4
sum is finite implies that the nk go to infinity as k goes to infinity. But, like
I said, we don’t have these tools yet. This is just a way to convince yourself
that the nk do in fact go to infinity. But, this way of thinking is very helpful
for convincing yourself that a result is true so that you may be able to prove
it. Yet another way of convincing yourself that the nk go to infinity is to
convince yourself that n1k have to go to 0 in order for Nk to converge (think
about why this is true).
At this point we know that Nk → L and that k < nk < nk+1 . To show
that L is the least upper bound, we will assume that there is an upper bound
less than L and show that we didn’t take the biggest step we could have
somewhere along the way. Let’s prove it.
Let B be an upper bound of S. Assume B < L (i.e., L is not the least
upper bound of S). Then, let δ = L − B > 0. By the Archemedian property
1
of the real numbers, there exists an n∗ ∈ N such that ∗ < δ. By the
n
1
convergence of Nk , there exists a K0 such that for all k ≥ K0 , |Nk − L| < ∗ .
n
1
1
At this point, we know that EVERY Nk is between L − ∗ and L + ∗ ,
n
n
1
provided k ≥ K0 . But, D = L − δ < L − ∗ . Therefore D is an upper bound
n
1
of S, and for every k ≥ K0 , Nk − ∗ > D. So, we get our contradiction if we
n
have a k ≥ K0 for which nk > n1∗ . (This is why the SIDETHOUGHT above
is true). Now, let k = max{n∗ , K0 }. Then, n∗ ≤ k + 1 and by the lemma,
1
k + 1 < nk+1 . Therefore, n∗ < nk+1 . But then, Nk − ∗ > D. This implies
n
1
∗
that Nk − ∗ is an upper bound for S, and n < nk+1 . This is a contradiction
n
1
is
to the fact that nk+1 is the smallest natural number such that Nk −
nk+1
an upper bound. Therefore, the assumption that B < L is false, i.e. B ≥ L
for all upper bounds B of S. Hence, (by definition!) L is the least upper
bound of S.
As promised, here is the proof of the lemma:
Proof of Lemma 2: This is pretty tedious. In order to simplify our lives,
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we first show that nk+1 ≥ nk . Then, we show that they can not be equal.
Finally, we show that k < nk . We do the first two by contradiction.
First, note that nk ≥ 2 for all k, since N −
1
is NOT an upper bound of
1
S! (This avoids technical difficulties later).
Assume nk+1 < nk . (This is the statement that on the k + 1st step we
went further than on the k th step). We know that each Nk is an upper bound
of S. We also know that the sequence Nk is decreasing. Thus,
Nk−1 −
1
nk+1
> Nk −
1
nk+1
= Nk+1 ,
1
i.e. Nk−1 − nk+1
is an upper bound of S, and nk+1 < nk . This contradicts the
fact that nk is the SMALLEST natural number n for which Nk−1 − n1 was an
upper bound of S. Hence, the assumption is false, and nk+1 ≥ nk for every k.
1
Assume that nk = nk+1 . Then, Nk+1 = Nk − nk+1
. But, Nk = Nk−1 − n1k .
Therefore,
1
1
−
.
Nk+1 = Nk−1 −
nk nk+1
However, nk+1 = nk . Therefore,
Nk+1 = Nk−1 −
2
nk
Now, note that since nk ≥ 2
2
1
≥
nk
nk − 1
we have the inequality
Nk+1 = Nk−1 −
2
1
≤ Nk−1 −
.
nk
nk − 1
But, this means that Nk−1 − nk1−1 is an upper bound of S, with nk − 1 < nk .
This contradicts the fact that nk was the smallest natural number n for
1
which Nk−1 − is an upper bound. Therefore, our assumption is invalid,
n
and nk 6= nk+1 for all k.
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This, together with the previous result gives that nk < nk+1 for all k.
Finally, the last little piece is easy. Since nk < nk+1 , we must have that
nk+1 ≥ nk + 1. This, with the fact that n1 > 1 implies that nk > k for all
k (simply by repeatedly adding 1 to both sides of the indequality, and using
the two inequalities given).
Hence, with the construction of Nk as in the proof with the corresponding
nk , we have
k < nk < nk+1
FOR ALL k
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