Dedekind Completeness and Limits

Limit of a Sequence

In the usual calculus materials, the limit of a sequence \(\left(a_n\right)\to L\) is defined by an analogous definition of the usual “$\epsilon$-$\delta$ definition” of the limit:

\begin{equation} \label{eqn:def-of-limit}\forall\epsilon>0\exists N\in\mathbb{N}\forall n\geq N;\ \vert a_n-L\vert <\epsilon. \end{equation}

This itself does not sound that odd as long as one is used to the $\epsilon$-$\delta$ definition of the limit. Following this definition, the limit $L=\lim_{n\to\infty}a_n$ is the “filler” of a logical statement \eqref{eqn:def-of-limit}.

On the other hand, the limit may be understood as a guess from the tendency of the sequence. So for instance, it is not hard to guess that

I would say that, this example suggests that finding a limit is an activity associated with our sense of quantities. And, if something is clinged with some ‘untold dependency to our sense,’ usually this boils down to an axiom that we no longer want to justify further.

Dedekind Completeness

One of the fundamental properties of the real number is so-called the completeness axiom:

If a nonempty subset $S\subset\mathbb{R}$ is bounded above, then it has the supremum $\sup S$.

(Dual Fact) If a nonempty subset $S\subset\mathbb{R}$ is bounded below, then it has the infimum $\inf S$.

If one allows $\sup S=\infty$ for $S$ unbounded above, and $\inf S=-\infty$ for $S$ unbounded below, then we can restate this axiom as

Any nonempty subset $S\subset\mathbb{R}$ of reals has the supremum $\sup S$.

(Dual Fact) Any nonempty subset $S\subset\mathbb{R}$ of reals has the infimum $\inf S$.

The fact that this completness is given as an axiom, pretty much indicates that there is no systematic way to find the supremum/infimum of a given set. (Apparently false, if one believes in the “Dedekind cut” construction of the reals. But I am demanding a method independent to the construction details here.)

More aspects of $\sup S$ and $\inf S$:

• These quantities are easily understood in the “precalculus-level,” where one can sketch the set $S$ on the real line, and ‘slide’ the upper/lower bounds to reach to the supremum/infimum.
• If $S$ is given by a monotonic sequence, listing the decimal expressions makes the guess for $\sup S$ or $\inf S$ quite well (as long as we list sufficiently many terms).
• Let $M<\infty$ (rspt. $m>-\infty$). Proving $M=\sup S$ (rspt. $m=\inf S$) is equivalent to, for any $\epsilon>0$ one has $x\in S$ in which $M\geq x>M-\epsilon$ (rspt. $m\leq x<m+\epsilon$).
  Thus suprema/infima does reflect an infinitesimal phenomena.

Collecting these, it seems like the completeness axiom is (1) associated with our visual understanding of the real line, while (2) very closely related to the infinitesimal, or limiting phenomena as well. Brought from the observations, I claim that the concept of the limit is built based on the completeness axiom.

As a support of the claim, let us see some “formula of limits” written in terms of superma and infima. They are particularily intriguing as they (superficially) excludes the $\epsilon$-$\delta$ or $\epsilon$-$N$ definition of limits.

From Completeness to Sequence Limits

To state how the limit of a sequence is found, the starting point will be the monotone convergence theorem, an equivalent formulation of the completeness axiom above.

Any increasing sequence \(\left(a_n\right)\) has the limit \(\sup a_n\) (possibly $=\infty$).

(Dual Fact) Any decreasing sequence \(\left(a_n\right)\) has the limit \(\inf a_n\) (possibly $=-\infty$).

Moreover, there is a generic way to generate an increasing or decreasing sequence, from the given sequence \(\left(a_n\right)\). Namely,

(If \(\left(a_n\right)\) is not bounded, then these sequences may take the values $\pm\infty$. We intentionally allow this.) There, $\left(b_k\right)$ gives a decreasing sequence while $\left(c_k\right)$ gives an increasing sequence, thus should have their respective limits $\inf b_k$ and $\sup c_k$.

These sequences are the ingredients of so-called the limit supremum (limsup) or limit infimum (liminf) of the sequence:

These quantities are relevant to the limit, as they are independent with any adjustment of $\left(a_n\right)$ for the first few terms. That is, $\limsup a_n$ and $\liminf a_n$ are “tail” quantities: they are independent to changing a finite number of terms in $a_n$.

If these limsup and liminf coincide, then one can say that the limit \(\lim_{n\to\infty}a_n\) exists and equals to that coinciding value. The converse is also the case.

(Proof) Suppose \(\limsup a_n=\liminf a_n=L\). If $L=\infty$, then for each $M>0$ we have \(c_N\) in which \(M<c_N\). If $n\geq N$ then \(c_N\leq a_n\) holds, thus \(M<a_n\) follows. This proves $\lim a_n=\infty$.

Likewise for $L=-\infty$, but focused on \(\left(b_k\right)\).

If $L$ is finite, then for any $\epsilon>0$ we have $K,M\in\mathbb{N}$ in which $L-\epsilon<c_M$ and $b_K<L+\epsilon$ holds. Let $N=\max(K,M)$. Then for any $n\geq N$, we have

\[L-\epsilon<c_M\leq c_N\leq a_n\leq b_N\leq b_K<L+\epsilon.\]

This verifies that $\vert a_n-L\vert <\epsilon$.

Conversely, if $\lim a_n=L$ (with $L$ finite), then for any $\epsilon>0$, one can choose $N\in\mathbb{N}$ in which all $n\geq N$ satisfies $L-\epsilon<a_n<L+\epsilon$. Because of that, we have $L-\epsilon\leq \inf_{n\geq N}a_n=c_N$ and $L+\epsilon\geq\sup_{n\geq N}a_n=b_N$. This entails,

\[L-\epsilon\leq c_N\leq\sup c_k=\liminf_{n\to\infty}a_n, \\ L+\epsilon\geq b_N\geq\inf b_k=\limsup_{n\to\infty}a_n.\]

That is, $L-\epsilon\leq\liminf a_n\leq\limsup a_n\leq L+\epsilon$ holds for all $\epsilon>0$. The only possible state for this to hold is $\liminf a_n=\limsup a_n=L$.

(Notice that we have used the ‘archimedean property,’ that the only number $x$ in which $\vert x\vert <\epsilon$ for all $\epsilon>0$ is $x=0$, crucially in the converse. The fact that $\liminf a_n\leq\limsup a_n$ is not heavily hard as long as one gets $c_\ell\leq b_k$ for all $k,\ell\geq 1$.)

Some example demos:

• If $a_n=(-1)^n/n$, then \(b_k=\sup_{n\geq k}a_n=\begin{cases} \frac{1}{k+1} & \text{ if }k\text{ is odd} \\ \frac{1}{k} & \text{ if }k\text{ is even}\end{cases}\), while \(c_k=\inf_{n\geq k}a_n=\begin{cases} -\frac{1}{k} & \text{ if }k\text{ is odd} \\ -\frac{1}{k+1} & \text{ if }k\text{ is even}\end{cases}\).
  There one sees that $b_k\colon\frac12,\frac12,\frac14,\frac14,\frac16,\frac16,\cdots$ has infimum 0, while $c_k\colon-1,-\frac13,-\frac13,-\frac15,-\frac15,\cdots$ has supremum 0. This tells $\limsup a_n=\liminf a_n=0$, thus the limit.
• If $a_n=\frac12(1+(-1)^n)\colon 0,1,0,1,0,1,\cdots$, then $b_k=\sup_{n\geq k}a_n=1$ while $c_k=\inf_{n\geq k}a_n=0$. This evidently tells $\limsup a_n=1\neq 0=\liminf a_n$, thus the limit fails to exist.

From Completeness to Function Limits

The above limsup and liminf for sequences are somewhat classical topics in analysis. But in fact, one can tweak the definitions and define the following limsup and liminf of functions:

Again, by the completeness axiom (if we admit $\pm\infty$) the quantities always exist. Note also that these quantities are independent to any tweak of values off from $a$. That is, even if I change values of $f(x)$ on \(\{x\mid\vert x-a\vert >\delta'\}\), I still have the same $\limsup_{x\to a}f(x)$, etc.

Furthermore, they coincide with the value $L$ iff $\lim_{x\to a}f(x)=L$.

(Proof) Suppose $\limsup_{x\to a}f(x)=\liminf_{x\to a}f(x)=L$. If $L$ is finite, then for each $\epsilon>0$ one has $\delta_1,\delta_2>0$ in which,

\[L-\epsilon<\inf\{f(x)\mid 0<\vert x-a\vert <\delta_2\}, \\ \sup\{f(x)\mid 0<\vert x-a\vert <\delta_1\} < L+\epsilon.\]

Set $\delta=\min(\delta_1,\delta_2)$, and let $x$ be any number such that $0<\vert x-a\vert <\delta$. Then we have

\[L-\epsilon < \inf_{0<\vert x-a\vert<\delta_1}f(x) \leq f(x) \leq \sup_{0<\vert x-a\vert<\delta_2} f(x)<L+\epsilon.\]

There $\vert f(x)-L\vert <\epsilon$ follows, and by this we check the usual `$\epsilon$-$\delta$ definition.’

The cases if $L=\pm\infty$ and the converse will be skipped. (Essentially same as the sequence cases.)

So for $f(x)=\sin(1/x)$ with $x\to 0$, we have, for any $\delta>0$,

thus the $\limsup_{x\to 0}f(x)=1$ and $\liminf_{x\to 0}f(x)=-1$ follows. Other than playing with the negation of $\epsilon$-$\delta$ definition, this check is much more straightforward to deal with!

Another example is $f(x)=x$ with $x\to a$. For any $\delta>0$, we have

Hence $\limsup_{x\to a}x=\inf_{\delta>0}(a+\delta)=a$, and likewise $\liminf_{x\to a}x=a$. Without any fancy $\epsilon$-$\delta$ type arguments, we thus have proven that $\lim_{x\to a}x=a$ ! This sort of mathematics is similarily applied for cases like $f(x)=x^2$, $x\to 1$, or $f(x)=1/x$, $x\to 2$, etc.

The matter becomes a little tricky for $f(x)=x\sin(1/x)$ with $x\to 0$. For any $\delta>0$, it is hard to estimate

exactly. However, as $-\vert x\vert\leq x\sin\frac1x \leq \vert x\vert$, we see that the suprema and infima of interest is estimated

This entails $0\leq\liminf f(x)\leq\limsup f(x)\leq 0$, thus the limit of interest. There, we see that we were simply lucky enough to find the exact limit value with this method (the squeeze theorem!).

Disclaimer

In case you view the “limsup-liminf definition” of limit,

We say $\lim_{x\to a}f(x)=L$ if $\limsup_{x\to a}f(x)=\liminf_{x\to a}f(x)=L$.

as a nice (replacing) definition, then I would note that this definition is (simply) losing too many things, compared with the standard $\epsilon$-$\delta$ definition:

• exercising local estimates of the error $\vert f(x)-L\vert$
• gadgets to prove limit theorems (E.g.: proving $\lim(f(x)\cdot g(x))=\lim f(x)\cdot\lim g(x)$ with the “limsup-liminf definition” is a mess.)
• generalizations of the concept towards metric or topological spaces (probably the most critical pitfall!)

and many more. So although I demo’ed that

1. suprema and infima are quite natural in accordance to the completeness axiom (and visual sketches),
2. formalizing the limit definitions in accordance to them may give an easier access to the concept of limits, and
3. the limit phenomena on the real line thus can be understood as a consequence of the completeness axiom,

I still believe in myself that the (modern standard) $\epsilon$-$\delta$ definition of limit should be kept and not to be replaced by the “limsup-liminf definition.”

Completely Regular Spaces (2022.01.18. appendix)

A comment from Junekey Jeon is that the “limsup-liminf definition” may be extended to completely regular spaces (also known Tychonoff spaces, or $T_{3\frac12}$ spaces).

Before talking anything further, first think of a sequence $\left(a_n\right)$ in $Y$, where $Y=\mathbb{R}^I$ for some set $I$. For instance, if $Y=\mathbb{C}$ is the complex plane, then we set \(I=\{1,2\}\). In that case, we think $\left(a_n\right)$ as a family of sequences $i\in I\mapsto\left(a^i_n\right)$. There, we say

$\left(a_n\right)\to L=\left(L^i\right)_{i\in I}$ if each $\left(a^i_n\right)\to L^i$, i.e., $\limsup a^i_n=\liminf a^i_n=L^i$ for all $i\in I$.

(In fact, this is a way how one characterizes the product topology on $Y=\mathbb{R}^I$. See Wiki:Convergence space.)

Now a completely regular space $X$ means, $X$ can be viewed as a subspace $X\subset\mathbb{R}^I$ for some set $I$ (See here). Thus theoretically, we can generalize the “limsup-liminf definition” to fairly general topological spaces. (For instance, you may view $X$ as a CW complex or a manifold.)

Even, this “theoretical throw-in” may be said more concretely. Put $I=C(X)$ for the set of all continuous functions $X\to\mathbb{R}$. Then we can view a completely regular space $X$ as $X\subset\mathbb{R}^{C(X)}$ by $x\mapsto\left(f(x)\right)_{f\in C(X)}$. Deciphering this encoding and tayloring this to a limit definition on $X$, we have:

$\left(a_n\right)\to L$ if for any continuous $f\colon X\to\mathbb{R}$, we have $\left(f(a_n)\right)\to f(L)$.

Update Log

• 220112: Created
• 220112: Feedbacks from Junekey Jeon applied
• 220118: Added a note on completely regular spaces