Proof Verification: Tao, Analysis I: Lemma 5.3.14.

Question

Preliminaries:

$\textbf{Lemma 5.3.14:}$Let $x\in\Bbb{R}$ such that $x\neq 0$. Then $x=LIM_{n\rightarrow\infty}$ for some Cauchy sequence $(a_n)_{n=1}^{\infty}$ which is bounded away from zero.

Proof:(by contradiction) Let $x\in\Bbb{R}$ such that $x\neq 0$. Then by definition 5.3.1., $x=LIM_{n\rightarrow\infty}a_n$ such that $(a_n)_{n=1}^{\infty}$ is a Cauchy sequence; That is \begin{align}\forall\epsilon\in\Bbb{Q^+}\exists N\in \Bbb{Z^+}\forall j,k\geq N(\vert a_j-a_k\vert\leq\epsilon)\end{align} and $(a_n)_{n=1}^{\infty}$ is also bounded. Suppose $x=LIM_{n\rightarrow\infty}a_n$ is not bounded away from zero. Thus, $\exists j\in\Bbb{N}$ such that $a_j=0$.

Consider $x'=LIM_{n\rightarrow\infty}a'_n$ where $\forall n\in\Bbb{N}(n\neq j\Leftrightarrow a_n=a'_n)$. Consequently due to definition 5.3.4. and Lemma 5.3.6.
\begin{align}x-x'=LIM_{n\rightarrow \infty}a_n - LIM_{n\rightarrow \infty}a'_n=LIM_{n\rightarrow \infty}(a_n-a'_n)\end{align} Thus, $\forall n\in\Bbb{N}(n\neq j\Leftrightarrow a_n-a'_n=0)$. Keep in mind that $x,x'\in\Bbb{R}$ and $\forall n\in\Bbb{N},a_n,a'_n\in\Bbb{Q^+}$. Since $a'_j\neq a_j=0$, then by trichotomy of rationals $a'_j>0$ or $a'_j<0$. Without loss of generality, suppose $a'_j>0$ then $\vert a_j-a'_j\vert =\vert 0-a'_j\vert =\vert -a'_j\vert=-(-a'_j)=a'_j$. Hence, $LIM_{n\rightarrow\infty}(a_n-a'_n)$ where $(a_n-a'_n)_{n=1}^{\infty}$ represents the sequence \begin{align}a_1-a'_1=0,...,a_{j-1}-a'_{j-1}=0,a_j-a'_j=a'_j,a_{j+1}-a'_{j+1}=0,0,...\end{align} Clearly, for all $\epsilon'\in\Bbb{Q^+}$ there exists $\overline{n}\in\Bbb{Z^+}$ such that $\forall k\geq\overline{n},\vert a_k-a'_k\vert\leq\epsilon'$. Finally consider $(x')^{-1}=LIM_{n\rightarrow\infty}(a'_n)^{-1}$ where $\forall n\in\Bbb{N},a'_n(a'_n)^{-1}=(a_n)^{-1}a_n=1$. By proposition 5.3.10., Multiplication of Real numbers is well-defined. Therefore, it must be that $x'(x')^{-1}=(x')^{-1}x'=(x')^{-1}x=x(x')^{-1}=1$. However, $\exists j\in\Bbb{N}$ such that $a_j=0$ thus $x$ is not invertible which yields the contradiction with proposition 5.3.10. Therefore $(a_n)_{n=1}^{\infty}$ must be bounded away from zero.

Answer 1

This line in the proof needs justification:

Suppose $x = LIM_{n \to \infty} a_n$ is not bounded away from zero. Thus, $\exists\ j \in \mathbb{N}$ such that $a_j = 0$.

This is because the contrapositive should read as follows:

If $x = LIM_{n \to \infty} a_n$ is not bounded away from zero, then for every rational number $c > 0$ there exists $n \geq 1$ such that $| a_n | < c$.

So, we need to rule out the case that infinitely many terms of the sequence satisfy this condition. That is, we need to show that if there is a subsequence $\{ a_{n_k} \}$ such that $|a_{n_k}| < 1/k$ for each $k \geq 1$, then we have a contradiction.

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Justification:

Let $x = LIM_{n \to \infty} a_n$ is a real number different from zero. Suppose that $\{ a_n \}$ is not bounded away from zero. So, for every rational number $c > 0$ there exists $n \geq 1$ such that $|a_n| < c$.

Suppose that $\{ a_{n_k} \}$ is a subsequence of $\{ a_n \}$ such that $|a_{n_k}| < 1/k$ for all $k \geq 1$. We know that for every rational $\epsilon > 0$ there exists $N \geq 1$ such that $|a_n - a_m| < \epsilon$ for all $n,m \geq N$. In particular, we can choose $N$ large enough so that we also have $|a_{n_k}| < \epsilon$ for all $k \geq N$.

Now, we will show that $\{ a_n \}$ is equivalent to $\{ 0 \}$, which is the additive identity in the set of real numbers. This will contradict the assumption that $x \neq 0$.

For every $m, k \geq 1$, we have that $$ |a_m - 0| = |a_m| = |a_m - a_{n_k} + a_{n_k}| \leq |a_m - a_{n_k}| + |a_{n_k}|. $$ So, for every $\epsilon > 0$, there exists $N \geq 1$ such that $$ |a_m| \leq |a_m - a_{n_k}| + |a_{n_k}| < \epsilon + \epsilon = 2 \epsilon $$ for all $m,k \geq N$. In particular, $$ |a_m| < 2 \epsilon $$ for all $m \geq N$. Hence, $\{ a_n \}$ is equivalent to $\{ 0 \}$, so $x = 0$, which is a contradiction.

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Thus, if for every $n \geq 1$ there exists $a_n$ such that $|a_n| < 1/n$, then there cannot be infinitely many such $a_n$'s. Let $a_{n_1},\dots,a_{n_M}$ be all such $a_n$'s. If they are all non-zero, choose $c = \min\{ |a_{n_i}| : 1 \leq i \leq M \} / 2 > 0$. This value of $c$ shows that $\{ a_n \}$ is bounded away from zero, which contradicts our initial assumption. So, $a_{n_i} = 0$ for at least one $i = 1,\dots,M$. From the previous discussion, we know that there can only be finitely many $a_n = 0$. In particular, there is no loss of generality in assuming that $a_{n_i} = 0$ for all $1 \leq i \leq M$.

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In your proof you also seem to be implicitly assuming that there is exactly one $a_j$ such that $a_j = 0$, whereas this need not be the case. But since there are only finitely many such $a_j$ (from the previous discussion), rectifying this should not be a problem.

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Lastly, the part following this sentence:

Clearly, for all $\epsilon'\in\Bbb{Q^+}$ there exists $\overline{n}\in\Bbb{Z^+}$ such that $\forall k\geq\overline{n},\vert a_k-a'_k\vert\leq\epsilon'$.

is incorrect. You have actually constructed a Cauchy sequence $\{ a'_n \}$ which is equivalent to $\{ a_n \}$ and is bounded away from zero. This shows what you need. This fact also shows that what you have is really a proof by construction and not a proof by contradiction.