I'm currently studying Tao's Analysis. I'm having difficult with proving Proposition 6.1.15, which says:

Proposition 6.1.15 (Formal limits are genuine limits.) Suppose that $(a_n)_{n=1}^{\infty}$ is a Cauchy sequence of rational numbers. Then $(a_n)_{n=1}^{\infty}$ converges to $\text{LIM}_{n\to\infty}a_n$, i.e. \begin{equation} \text{LIM}_{n\to\infty}\,a_n = \lim_{n\to\infty}\,a_n. \end{equation}

Here are some clarification for the unusual notations. The author use $\text{LIM}_{n\to\infty}\,a_n$ to define a real number, where $a_n$ is a Cauchy sequence. And formal limit means the limit of a Cauchy sequence. Here is the definition:

Definition 5.3.1 (real numbers). A real number is defined to be an object of the form $\text{LIM}_{n\to\infty}\,a_n$, where $(a_n)_{n=1}^{\infty}$ is a Cauchy sequence of rational numbers. Two real numbers $\text{LIM}_{n\to\infty}\,a_n$ and $\text{LIM}_{n\to\infty}\,b_n$ are said to be equal iff $(a_n)_{n=1}^{\infty}$ and $(b_n)_{n=1}^{\infty}$ are equivalent Cauchy sequences. The set of all real numbers is denoted $\mathbf{R}$.

Tao gives the following outline:

Let $(a_n)_{n=m}^{\infty}$ be a Cauchy sequence of rationals, and writing $L:= \text{LIM}_{n\to\infty}\,a_n$. We have to show that $(a_n)_{n=m}^{\infty}$ converges to $L$. Let $\epsilon \gt 0$. Assume for sake of contradiction that sequence $a_n$ is not eventually $\epsilon$-close to $L$. Use this, and the fact that $(a_n)_{n=m}^{\infty}$ is Cauchy, to show that there is an $N\ge m$ such that either $a_n \gt L + \epsilon /2$ for all $n \ge N$, or $a_n \lt L - \epsilon/2$ for all $n \ge N$. Then use Exercise 5.4.8(which says, if $a_n \ge x$ for all $n \ge 1$, then the formal limit $\text{LIM}_{n\to\infty}\,a_n\ge x$).

My Attempt:

Let $\epsilon > 0$. Assume $a_n$ is not eventually $\epsilon$-close to $L$ . Since $a_n$ is Cauchy sequence, there exists $N\ge m$ such that $| a_n - a_{n'} | \lt \epsilon$, for every $n$, $n' \ge N$. Since by assumption, $a_n$ is not eventually $\epsilon$-close to $L$, we know for this $N$, there exists $n_1 \ge N$ such that \begin{equation} | a_{n_1} - L | \ge \epsilon. \end{equation} Now, for $n_1$, there exists $n_2 \ge n_1 \ge N$ such that \begin{equation} | a_{n_2} - L | \ge \epsilon. \end{equation} But $| a_{n_1} - a_{n_2} | < \epsilon$, which implies either $a_{n_1}$ and $a_{n_2}$ are both greater than or equal to $L+\epsilon$ or both less than or equal to $L-\epsilon$ (i.e. they are on the same side with respect to $L$). Repeated the process, we can obtain $a_{n_1}, a_{n_2}, \cdots, a_{n_k}$, which are all great than or equal to $L+\epsilon$ or less than or equal to $L-\epsilon$.

Up to now, we only get a subsequence of $a_n$, which satisfies the condition $a_{n_k} \ge L + \epsilon \gt L + \epsilon/2$(or the other direction). Here is where I stuck, how to show there exists $N$ such that $a_n \gt L+\epsilon/2$ for all $n\ge N$?

Can you give me a hint? Thanks for you time!

  • $\begingroup$ "LIM", and the notion of a "formal limit", are not standard terminology. Probably it's specific to your textbook. Could you provide the definitions? $\endgroup$
    – Jack M
    Mar 29, 2017 at 6:33
  • $\begingroup$ @JackM Sure, I have added them now. $\endgroup$
    – Han Tang
    Mar 29, 2017 at 6:50

1 Answer 1


Ok, let me answer my own question here. If it is incorrect, feel free to point it to me! I will follow the author's outline, here is my proof:

We need to find a $N \ge m$ such that either $a_n \gt L + \epsilon/2$ for all $n \ge N$, or $a_n \lt L - \epsilon/2$ for all $n \ge N$. Let $\epsilon \gt 0$. Assume for the sake of contradiction that $a_n$ is not eventually $\epsilon$-close to $L$. Since $a_n$ is a Cauchy sequence, we know for $\epsilon/2$, there exists $N' \ge m$ such that \begin{equation} | a_n - a_{n'} | \le \epsilon/2, \; \text{for every}\, n,n' \ge N'. \end{equation} Because $a_n$ is not eventually $\epsilon$-close to L, we know for $N'$, there exists $N\ge N'$ such that \begin{equation} | a_N - L | \gt \epsilon. \end{equation} So, $a_N \gt L + \epsilon$ or $a_N \lt L - \epsilon$. First, consider the case $a_N \gt L + \epsilon$. Since, $N \ge N'$, we have \begin{equation} | a_n - a_N | \le \epsilon/2, \; \text{for every} \; n \ge N \ge N', \end{equation} which implies $a_n \ge a_N - \epsilon/2 \ge L + \epsilon - \epsilon/2 = L + \epsilon/2$, for every $n \ge N$. But that implies $L = \text{LIM}_{n\to\infty}{a_n}\ge L + \epsilon/2$, which is a contradiction.

Similarly, consider the case $a_N \lt L - \epsilon$, we can finally get $a_n < L -\epsilon/2$, for $n\ge N$, which will also lead to a contradiction.

Therefore, $a_n$ must be eventually $\epsilon$-close to $L$ for every $\epsilon \gt 0$, i.e. $L = \lim_{n\to\infty}a_n$, as desired.


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