Equivalence between two definitions for the limit of a sequence

Question

Bartle, in "The Elements of Integration and Lebesgue Measure", defines the limit of a sequence as it follows:

However, Rudin, in "Principles of Mathematical Analysis", defines it otherwise:

I would like to prove both definitions to be equivalent.

Answer 1

Here is a quick attempt at a proof, feel free to correct anything. The notation is especially suspect, but at least this should give you some ideas.

Assuming $x_n \in \mathbb R$ definition for limit superior $\beta_0$ is:

$$\forall \epsilon >0(\exists N\forall n (n\geq N \rightarrow x_n<\beta_1 +\epsilon)\land \exists N\exists n(N\geq n \rightarrow x_n>\beta_1 -\epsilon)))$$

and limit inferior $\beta_1$:

$$\forall \epsilon >0(\exists N\forall n( n\geq N \rightarrow x_n>\beta_1 -\epsilon)\land \exists N\exists n(N\geq n \rightarrow x_n<\beta_1 +\epsilon)))$$

The first definition would be those two combined with a conjunction with an added conjunction for $\beta_0 = \beta_1$. Now we set $\beta_0 =\beta_1 $ (equivalent to adding an and symbol after both the implications - not going to prove this as it seems difficult), and then solve the equations after the implications so that epsilon is on the RH side:

$$\forall \epsilon >0(\exists N\forall n(n\geq N \rightarrow (x_n-\beta_1< \epsilon \land -x_n+\beta_1 < \epsilon) \\ \land \exists N\exists n(N\geq n \rightarrow (-x_n+\beta_1< \epsilon\land x_n-\beta_1<\epsilon)))$$

Now note that the part after both implications is the definition of absolute value.

$$\forall \epsilon >0(\exists N\forall n(n\geq N \rightarrow (|x_n-\beta_1|< \epsilon) \\ \land \exists N\exists n(N\geq n \rightarrow |x_n-\beta_1|< \epsilon))$$

and finally we arrive at the 2nd definition, after setting $B_1$ to L, and noting that the 2nd row is always true if the first is (if something exists for all, surely it exists for specific case). Thus:

$$\forall \epsilon>0 \space\exists N \forall n(n\geq N \rightarrow |x_n-L|<\epsilon)$$

Where if the definition holds $L$ is the limit. We have arrived at the 2nd definition.

Answer 2

First, it's important to note that these definitions are equivalent exclusively in the context of $\mathbb{R}$. To show the equivalence, we must show two things:

(I) If $\limsup x_n = \liminf x_n = L$, then $L = \lim x_n$.

(II) If $\lim x_n = L$, then $\liminf x_n = \limsup x_n = L$.

The latter should be easier to show. Suppose $\lim x_n = L$ (in the sense of Rudin). I'll prove the $\limsup x_n \equiv \inf_{m} ( \sup_{n \geq m} x_n ) = L$. To do so, I'll use the following lemma: If $S = \limsup x_n$, then there exists a sequence $n_1 < n_2 < \cdots$ such that $\lim_k x_{n_k} = S$. In other words, there exists a subsequence of $(x_n)$ that converges (in the sense of Rudin) to the limit superior. To see this, we'll construct the $(n_k)$ as follows. But before we do, observe that the sequence $( \sup_{n \geq 1} x_n, \sup_{n \geq 2} x_n, \ldots)$ is decreasing, so $\inf_{m \geq k} \sup_{n \geq n} x_n = \inf_{m \geq 1} \sup_{n \geq n} x_n$.

With this fact in mind, we can prove the lemma. First, let $n_1$ be such that \begin{align*} x_{n_1} & < (\inf_{m \geq 1} \sup_{n \geq m} x_{n}) + 1 \\ & = S + 1 . \end{align*} Now for the rest, assuming $n_{k}$ has been constructed, let $n_{k + 1} > n_{k}$ such that \begin{align*} x_{n_{k + 1}} & < (\inf_{m \geq n_k + 1} \sup_{n \geq m} x_{n}) + (k + 1)^{-1} \\ & = S + (k + 1)^{-1} . \end{align*} This sequence clearly converges (in the sense of Rudin) to $S$. In particular, $|x_{n_k} - S| < 1/k$.

Now let's take stock of what we've shown. We've shown that any sequence has a subsequence convergent to its limit superior. Now we claim that in the case where the sequence has a Rudin limit $L$, this $L$ is the limit superior. To see this, fix $\epsilon > 0$, and let $1/K < \epsilon$. By the fact that $(x_n)$ is Rudin-convergent, we know that there exists an $N$ such that if $n \geq N$, then $|x_n - L| < \frac{1}{2K}$. Now choose $n_{k}$ such that $k \geq 2K$ and $n_{k} \geq N$. Then \begin{align*} |L - S| & \leq |L - x_{n_k}| + |x_{n_k} - S| \\ & < \frac{1}{2K} + \frac{1}{2K} \\ & = 1/K \\ & < \epsilon . \end{align*}

Now notice we just chose an arbitrary $\epsilon > 0$. So if $L \neq S$, then we would have been unable to do the above if, say, $\epsilon = |L - S|$. Thus this implies $L = S$, i.e. that the Rudin limit (when it exists) coincides with the limit superior. An almost identical argument will show the Rudin limit is equal to the limit inferior. This proves (II).

Now we need to prove the other way, (I). So let $(x_n)$ be a sequence such that $\liminf x_n = \limsup x_n = L$. We wanna show that $L$ is also the Rudin limit of $(x_n)$. This time we'll prove it by contradiction. Suppose to the contrary that $L$ is not the Rudin limit of $(x_n)$. Now the Rudin definition of a limit is as follows: For every $\epsilon > 0$, there exists $N$ such that if $n \geq N$, then $|x_n - L| < \epsilon$. So the negation of this would be that for some $\epsilon > 0$, there does not exist $N$ such that if $n \geq N$, then $|x_n - L| < \epsilon$. Here's an equivalent version of that, and the version we'll use:

For some $\epsilon > 0$ (call it $\epsilon_0$), it is the case that for every $N \in \mathbb{N}$, there exists $n \geq N$ such that $|x_n - L| \geq \epsilon_0$.

So we're gonna assume for contradiction that that's the case. Construct a sequence $n_1 < n_2 < \cdots$ as follows: Let $n_1$ such that $|x_n - L| \geq \epsilon_0$. Now assuming $n_k$ has been constructed, let $n_{k + 1} \geq n_k + 1$ such that $| x_{n_{k + 1}} - L | \geq \epsilon_0$.

Now note that trivially, $x_{n_k} \neq L$. So either infinitely many terms of $x_{n_k}$ are greater than $L$, or infinitely many terms are less than $L$. In fact it's possible both are true, but we only need that at least one of these are the case. We may assume WLOG it's the former. So we have shown the following:

For all $m \in \mathbb{N}$, there exists $n \geq m$ such that $x_n \geq L + \epsilon_0$.

As such, we have $$\sup_{n \geq m} x_n \geq L + \epsilon_0$$ for all $m$, from which it follows that $$\inf_{m} \sup_{n \geq m} x_n \geq L + \epsilon_0 .$$ But this is a contradiction, as $\inf_{m} \sup_{n \geq m} x_n = \limsup x_n = L$, so we've just "proven" $L > L$. Thus we conclude that our assumption "$L$ is not the Rudin limit of $(x_n)$" was in fact false. This proves (I).

Now having proven the equivalence in $\mathbb{R}$, it's worth pausing a moment to consider what these definitions mean. Bartle's definition is in terms of order relations, with the limits superior and inferior defined in terms of elements of $\mathbb{R}$ being greater or less than each other. This definition is not fit to generalize to, say, a definition of convergence in the plane. Suppose I had a sequence of points in the complex plane and wanted to talk about the limit of that sequence. How would I generalize Bartle's definition? What does it mean for one point in $\mathbb{C}$ to be "greater than" or "less than" another?

Rudin' definition, on the other hand, is more "geometric", drawing attention to the concept of how "close" points are to each other. This is made possible by referring to the natural metric structure on $\mathbb{R}$. Rudin's definition is such that we could talk about, say, the limit of a sequence of points in the plane. Moreover, Rudin's definition is more fit to generalize to a definition of convergent sequences of functions. Bartle's definition will be able to capture only a few types of convergence, and it'll turn out that these types of convergence are often not the kind we're really interested in. Metric convergence is generally the more interesting and the more useful concept.