Math 316, Intro to Analysis
Limits superior and inferior.
We’ve proven some facts:
(1) A sequence converges if and only if it has exactly one limit point
(2) (Bolzano Weierstrass theorem) A bounded sequence has at least one limit point
Today we study the set of all limit points of a sequence.
Definition 1. Let an be a bounded sequence. the limit superior of an , lim sup(an ) = liman is
defined to be
The limit inferior of an , lim inf(an ) = liman is defined to be
This definition requires that the set of limit points of an be
and
. Why should this be the case?
For clarification: If I tell you that an is a sequence, that ank is a convergent subsequence
and an → a then how do a, lim inf(an ) and lim sup(an ) compare?
Proposition 2. A bounded sequence an converges if and only if lim sup(an ) = lim inf(an )
Proof. Let an be a bounded sequence.
=⇒
Suppose that an converges. Let a = lim(an ). We proved in the lecture on subsequences that
.
this means that a is the only limit point. Thus, {limit points of an } =
By definition, then
lim sup(an ) = sup
=
While
lim inf(an ) = inf
=
In particular,
lim sup(an )
lim inf(an )
⇐=
Suppose that lim sup(an ) = lim inf(an )and let a = lim sup(an ) = lim inf(an ). Let α be any
limit point of an .
Suppose that an does not converge to a, but negating the definition of convergence we see
that there exists an > 0 such that
In particular there are infinitely many n ∈ N such that |an −a|
. Let nk be a sequence
listing these n.
Thus, for all n |ank − a|
. Since an is bounded, ank is bounded. Thus, there is a
convergent sub-sub-sequence ankl Let b be its limit. Since
|ank − a|
1
,
2
it follows from limit laws that
|b − a|
and in particular a
b.
b is a sub sequential limit for an . Since a = lim sup(an ) is the supremum of
a
b. Since a = lim inf(an ) is the infemum of
b. Thus, a
b and a
b, a contradiction.
a
Thus, an must converge to a. In particular an is convergent.
,
Here is a useful alternative characterization of the limits superior and inferior.
Theorem 3. For a bounded sequence an ,
(1) L = lim sup(an ) if and only if for all > 0 there are infinitely many n with an ∈
(L − , L + ) and only finitely many with an > L + (2) L = lim inf(an ) if and only if for all > 0 there are infinitely many n with an ∈
(L − , L + ) and only finitely many with an < L − Proof. I will only do the proof of the first claim. The second proof is identical.
Suppose that L = lim sup(an ) then L is the supremum of all limit points of an .
Thus, for all > 0 there is a limit point a0 of an with
L − < a0 ≤ L.
2
Since a0 is a limit point of an , for infinitely many n
a0 − < an < a0 +
2
2
Combining these two inequalities, we see that for infinitely many n,
L − = L − − < a0 − < an < a0 + ≤ a0 + < L + 2 2
2
2
so that an ∈ (L − , L + ), as we claimed.
We still need to show that an > L + for only finitely many . Suppose that this was not
the case, that an > L + for infinitely many n. Then we can find a subsequence ank such
that ank > L + . But any subsequence has a convergent subsequence, so that there is a
convergent subsequence ankj with ankj → a (Notice that this means a is a limit point of an )
and ankj > L + . But then a ≥ L + > L.
But then we have a limit point a with a > L. L is not an upper bound on the set of limit
points. This completes half of the proof.
Conversely, suppose that for all > 0 there are infinitely many n with an ∈ (L − , L + )
and only finitely many with an > L + . We will show that L = lim sup(an ).
Let a be any limit point of an so that there is a subsequence ank converging to a. But for
all > 0, there con be only finitely many k with ank > L + , so that a ≤ L + . Since this is
true for all > 0 it follows that a ≤ L. L is an upper bound on the set of all limit points.
We now need to show that for all > 0 there is a limit point a0 with L − < a0 . Since for
all δ > 0 there are infinitely many n with an ∈ (L − δ, L + δ), it follows that L is a limit point
of an , but obviously L − < L. Thus L is the supremum of the set of all limit points.
The following corollaries have proofs which are good exercises
Corollary 4. lim sup an and lim inf(an ) are limit points of an .
Proof. Does part of the alternate characterization make you think of what out means to be
a limit point?
3
Theorem 5. For any bounded sequences an and bn ,
(1) lim sup(an + bn ) ≤ lim sup(an ) + lim sup(bn )
(2) lim inf(an + bn ) ≥ lim inf(an ) + lim inf(bn )
Proof. We have proven that lim sup(an + bn ) is a limit point of (an + bn ) so that there is a
subsequence with (ank + bnk ) → lim sup(an + bn ). The sequences (ank ) might not actually
converge, but it has a convergent subsequence (ankl ) → a. a is a limit point of an so that
a ≤ lim sup(an ).
Now bnkl = (ankl + bnkl ) − (ank ) converges to b := lim sup(an + bn ) − a. Since b is a limti
point of bn , b ≤ lim sup(bn ).
Finally, lim sup(an + bn ) = a + b ≤ lim sup(an ) + lim sup(bn ).
Why isn’t this an equal sign? Take an = (−1)n and bn = (−1)n+1 . Compute
(1) lim sup(an )
(2) lim sup(bn )
(3) an + bn
(4) lim sup(an + bn )
(5) Is the claim “If an and bn are sequences then lim sup(an + bn ) = lim sup(an ) +
lim sup(bn )” true of false?