limits superior and inferior

Section 2.5 limits superior and inferior

So in our quest to discover partial converses to Theorem 2.1.14, we’ve found the Monotone Sequence Theorem and the Bolzano-Weierstraß Theorem. There’s one other way to go: we could relax our notion of limit.

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Definition 2.5.1.

We call the number \(\alpha\) an accumulation point for the sequence \(X=\left(x_n\right)_{n\in\mathbb{N}}\) if, for any \(\epsilon\gt 0\text{,}\) there are infinitely many terms of \(X\) within \(\epsilon\) of \(\alpha\text{.}\) That is, \(\alpha\) is an accumulation point for \(X\) if:

\begin{equation*} \forall \epsilon\gt 0, \forall K\in\mathbb{N}, \exists n\gt K: \lvert x_n-\alpha\rvert\lt \epsilon\ \ . \end{equation*}

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The set of all accumulation points for \(X\) is denoted \(\operatorname{acc}(X)\text{.}\)

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What does this have to do with limits?

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Proposition 2.5.2.

\(\alpha\in \operatorname{acc}(X)\) if and only if there is a subsequence of \(X\) which converges to \(\alpha\text{.}\)

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Proof.

\(\Rightarrow\).

Given that \(\alpha\in \operatorname{acc}(X)\text{,}\) we need to construct a subsequence, which means we need to choose an increasing list of indices \(n_k\text{.}\)

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First let’s pick \(n_1\text{.}\) By the asumption that \(\alpha\in\operatorname{acc}(X)\text{,}\) there is some \(n\gt 1\) with \(\lvert x_{n}-\alpha\rvert\lt 1\text{.}\) Call this \(n_1\text{.}\)

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To pick \(n_2\text{,}\) use \(\epsilon=\frac{1}{2}\) in the definition of accumulation point. Then there is some \(n\gt n_1\) with \(\lvert x_n-\alpha\rvert\lt \frac{1}{2}\text{.}\) Call this \(n_2\text{.}\)

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Now, assuming we’ve already picked \(n_1\lt n_2\lt\cdots\lt n_k\) so that \(\lvert x_{n_j}-\alpha\rvert\lt \frac{1}{j}\text{,}\) we can apply the definition of accumulation point with \(\epsilon=\frac{1}{k+1}\) and \(K=n_k\) to find some \(n\gt n_k\) with \(\lvert x_n-\alpha\rvert\lt \frac{1}{k+1}\text{.}\) Call this \(n_{k+1}\text{.}\)

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In this way, we obtain a subsequence \(\left(x_{n_k}\right)_{k\in\mathbb{N}}\) with \(\lvert x_{n_k}-\alpha\rvert\lt \frac{1}{k}\text{,}\) which is to say

\begin{equation*} -\frac{1}{k}\lt x_{n_k}-\alpha \lt \frac{1}{k} \end{equation*}

By Two Officers and a Drunk, this means that \(x_{n_k}\to\alpha\text{.}\)

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\(\Leftarrow\).

Now, let’s say we have a subsequence \(\left(x_{n_k}\right)_{k\in\mathbb{N}}\) with \(x_{n_k}\to\alpha\text{.}\) We’ll verify the definition of accumulation point holds for \(\alpha\text{.}\) To this end, consider a given \(\epsilon\gt 0\text{.}\) By the definition of convergence, \(\exists K\in\mathbb{N}\) so that \(k\gt K\) guarantees \(\lvert x_{n_k}-\alpha\rvert\lt \epsilon\text{.}\) Observe that there are infinitely many such indices \(n_k\text{.}\)

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Example 2.5.3.

If \(x_n\to L\text{,}\) then \(\operatorname{acc}(X)=\{L\}\text{.}\)

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\(\operatorname{acc}\left((-1)^n\right)_{n\in\mathbb{N}}=\{-1,1\}\text{.}\)

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\(\operatorname{acc}\left( (-1)^n+\frac{1}{n}\right)_{n\in\mathbb{N}}=\{-1,1\}\text{.}\)

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For the sequence \(x_n=\sin(n)\text{,}\) \(\operatorname{acc}(X)=[-1,1]\text{.}\)

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Checkpoint 2.5.4.

We haven’t talked about cardinality in these notes, but a fact you ought to know is this:

The rational numbers \(\mathbb{Q}\) are countable. That is, there is a bijection between \(\mathbb{N}\) and \(\mathbb{Q}\text{.}\)
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In particular, there is an enumeration function \(\Phi:\mathbb{N}\to\mathbb{Q}\) which counts out each of the rational numbers exactly once.

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But an enumeration function is nothing more than a sequence.

In fact, Georg Cantor’s first published definition of a countable set was a set that can be written as the image of a sequence.

That is, there is a sequence \((q_n)_{n\in\mathbb{N}}\) for which each \(q_n\in\mathbb{Q}\text{,}\) and each rational number is \(q_n\) for a unique \(n\text{.}\)

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Compute \(\operatorname{acc}\left(q_n\right)_{n\in\mathbb{N}}\text{.}\)

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Aside

Georg Cantor was a mathematician who did foundational work on set theory in the late nineteeth century. Unlike Cauchy and Weierstraß, whose ideas had clear ancestors, Cantor appears to have genuinely come up with the notions of bijection, countability, and the existence of more than one kind of infinity completely on his own. This last idea, when it was introduced, was quite controversial. Among its detractors was Kronecker, who used his considerable influence within the mathematical community to make Cantor’s professional life quite difficult. The philosophical (even theological) implications of multiple infinities were seen as heretical. Cantor himself thought that God had given him the idea, and that it established the truth of pantheism. It did not help matters that Cantor’s wife, Vally Guttmann, was Jewish in a deeply anti-Semitic time and place.

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Ultimately, Cantor spent the last fifteen years of his life in and out of mental institutions, broken by the failure of his ideas to gain acceptance by some of the giants of the mathematical research community.

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Checkpoint 2.5.5.

The first example Example 2.5.3 says: A convergent sequence has exactly one accumulation point..

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Prove the converse: if \(X\) is bounded and \(\operatorname{acc}(X)=\{L\}\) is a singleton, then \(X\to L\text{.}\)

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Checkpoint 2.5.6.

Give an example of a sequence \(X\) which has exactly one accumulation point but is not convergent.

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So counting the number of accumulation points is one way to check convergence.

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Definition 2.5.7.

The limit superior of a bounded sequence \(X\) is

\begin{equation*} \limsup X=\sup\operatorname{acc}(X)\ \ . \end{equation*}

The limit inferior of a bounded sequence is

\begin{equation*} \liminf X=\inf\operatorname{acc}(X)\ \ . \end{equation*}

If \(X\) is not bounded above, we write \(\limsup X=\infty\text{.}\) If \(X\) is not bounded below, we write \(\liminf X=-\infty\text{.}\)

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Proposition 2.5.8.

A sequence \(X\) converges if and only if \(\limsup X=\liminf X\text{.}\)

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Proof.

This is just Checkpoint 2.5.5, restated using the \(\limsup\) and \(\liminf\) terminology.

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It can be a bit tricky to compute \(\limsup\) and \(\liminf\) directly -- you need to first find the accumulation points, and then find the supremum and infimum of that set. You might also wonder what’s with the notation \(\limsup\) and \(\liminf\text{.}\)

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Checkpoint 2.5.9.

Given a sequence \(X=\left(x_n\right)_{n\in\mathbb{N}}\text{,}\) for each \(k\in \mathbb{N}\text{,}\) consider the set \(T_k=\left\{x_n\middle\lvert n\gt k\right\}\text{.}\) Observe that \(T_{k+1}\subseteq T_k\text{.}\)

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The sequences \(I_k=\inf T_k\) and \(S_k=\sup T_k\) are monotone. Which flavor of monotone?

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If \(X\) is bounded, then these sequences are bounded. So they converge -- to what?

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Checkpoint 2.5.9 concerns a sensible interpretation of the notation \(\lim\ \sup\text{:}\) the limit of some suprema. That’s not identical to our notation for limit superior, which is \(\limsup\text{.}\)

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Theorem 2.5.10.

For a bounded sequence \(X\text{,}\)

\begin{equation*} \limsup X= \lim S_k=\lim(\sup T_k)=\inf\left\{S_k\middle\vert k\in \mathbb{N}\right\} \end{equation*}

and

\begin{equation*} \liminf X= \lim I_k=\lim(\inf T_k)=\sup\left\{I_k\middle\vert k\in \mathbb{N}\right\} \end{equation*}

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Proposition 2.5.11.

The limits superior and inferior of a bounded sequence are accumulation points for that sequence.

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Proof.

We’ll show that \(\limsup X\in\operatorname{acc}(X)\text{.}\)

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Our job is to construct a subsequence of \(X\) that converges to \(\limsup X\text{.}\)

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To find \(n_1\text{,}\) observe that using \(\epsilon=\frac{1}{2}\) in the definition of \(S_k\to\limsup X\text{,}\) there is some \(K_1\in\mathbb{N}\) so that \(r\gt K_1\) guarantees \(\lvert S_r-\limsup X\rvert\lt \frac{1}{4}\text{.}\) Pick such an \(r\text{,}\) say \(r_1=K_1+1\text{.}\) By the \(\epsilon\)-characterization of supremum, there is \(x_p\in T_{r_1}\) with \(S_{r_1}-x_p\lt \frac{1}{4}\text{.}\) Use this \(p\) as \(n_1\text{.}\) Observe that \(n_1\gt r_1\text{.}\) By the triangle inequality, \(\lvert x_{n_1}-\limsup X\rvert \lt \frac{1}{2}\text{.}\)

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To pick \(n_2\text{,}\) there is some \(K_2\in \mathbb{N}\) so that \(r\gt K_2\) guarantees \(\lvert S_r-X\rvert\lt \frac{1}{8}\text{.}\) Pick such an \(r\text{,}\) say \(r_2=n_1+K_2+1\text{.}\) By the \(\epsilon\)-characterization of supremum, there is \(x_p\in T_{r_2}\) with \(S_{r_2}-x_p\lt\frac{1}{8}\text{.}\) Call this \(p\) \(n_2\text{.}\) Observe that \(n_2\gt r_2\gt n_1\text{,}\) and \(n_2\gt K_2\text{,}\) so by the triangle inequality we’ve found \(n_2\gt n_1\) with \(\lvert x_{n_2}-\limsup X\rvert\lt \frac{1}{4}\text{.}\)

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Lather, rinse, and repeat to find \(n_1\lt n_2\lt n_3\lt \cdots\) so that \(\lvert x_{n_k}-\limsup X\rvert\lt \frac{1}{2^k}\text{.}\)

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Then by Two Officers and a Drunk, \(x_{n_k}\to\limsup X\) as required.

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The proof for \(\liminf X\) is similar.

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Checkpoint 2.5.12.

Could there be a bounded sequence whose set of accumulation points is \((-2,2)\text{?}\)

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Checkpoint 2.5.13.

Proposition 2.5.11 and its proof can be adapted to the more general statement:

If \(\alpha_n\) is a sequence of accumulation points for the sequence \(X\text{,}\) and \(\alpha_n\to L\text{,}\) then \(L\) is an accumulation point for \(X\text{.}\)
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Remark 2.5.14.

Checkpoint 2.5.13 shows that not every set could be \(\operatorname{acc}(X)\) for some \(X\text{.}\) More of this sort of thing shortly!

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