Prove that every Cauchy sequence in $\mathbb{R}$ converges.
proof: Let $A_n$ be a closed and bounded sequence. Then there exists an interval $[a_1,b_1]$ such that $a_1\leq A_n \leq b_1$ $\forall n$. Consider, $$I_1 = [a_1,(a_1 + b_1)/2] + [(a_1 + b_1)/2,b_1]$$ at least one has infinitely many elements of $A_n$. Pick one, and call this $I_2$. Split $I_2$ in half, and find $I_3$ with infinitely many points, proceed. (At the $jth$ step, we have $I_j$ containing infinitely many elements). Now, let's define a subsequence $a_{j_{k}}$, by letting it be the first element of the sequence $a_{j_{1}},\ldots,a_{j_{k-1}}$ which is an element of $I_j$. We claim that $a_{j_{k}}$ is a Cauchy sequence. Let's pick $k,l > N$, then both $a_{j_{k}}$ and $a_{j_{l}}$ lie in the interval $I_N$ which has length $(b-a)/(2^{n-1})$. Thus, $$|a_{j_{k}} - a_{j_{l}}| \leq (b-a)/(2^{n-1})$$ We can make the right hand side arbitrarily small by making $N$ sufficiently large. Thus we have shown that the subsequence is a Cauchy sequence and hence convergent.
I am not sure if I am right, any suggestion would be greatly appreciated.
$\endgroup$ 52 Answers
$\begingroup$It seems to me that you're trying to prove that every bounded sequence is $\mathbb R^\mathbb N$ has a convergent subsequence.
Note that by using this result we can easily prove that a real Cauchy sequence is convergent.
Proof 1: that every bounded sequence is $\mathbb R^\mathbb N$ has a convergent subsequence.
Let $(a_n)_{n\in \mathbb N} \in \mathbb R^\mathbb N$ be a bounded sequence. We suppose without loss of generality that it is bounded by $0$ and $1$.
We will now construct a convergent subsequence by defining a strictly increasing application $\varphi$ such that $(a_{\varphi(n)})_{n\in\mathbb N}$ is a convergent sequence.
Let $u_0=0$, $v_0=1$ and $w_0=\displaystyle\frac{u_0+v_0}{2}$
Let $A_0=\{n\in \mathbb N | u_0\leq a_n\leq w_0 \}$ and $B_0=\{n\in \mathbb N | w_0\leq a_n\leq v_0 \}$
One of them must be infinite, since their union is infinite:
If $A_0$ is infinite, we set $u_1=u_0$, $v_1=w_0$ and $w_1=\displaystyle\frac{u_1+v_1}{2}$.
Otherwise, we set $u_1=w_0$, $v_1=v_0$ and $w_1=\displaystyle\frac{u_1+v_1}{2}$.
We define $A_1$ and $B_1$ similarly, by using $u_1$, $v_1$ and $w_1$ instead of $u_0$, $v_0$ and $w_0$.
We define $u_2$, $v_2$, $w_2$, using the same method.
By repeating this process, we get two sequences $(u_n)_{n\in \mathbb N}$ and $(u_n)_{n\in \mathbb N}$ such that: $(u_n)_{n\in \mathbb N}$ is increasing, $(v_n)_{n\in \mathbb N}$ is decreasing, and the set $\{n\in \mathbb N|u_p\leq a_n\leq v_p\}$ is infinite for all $p\in \mathbb N$.
Since they are both bounded, then they both converge to the same limit $l$ (their difference goes to $0$ since it gets divided by $2$ after every step.)
We will now define our function $\varphi$ by recurrence.
The set $\{n\in \mathbb N|u_0\leq a_n\leq v_0\}$ is infinite. We choose any element of that set as $\varphi(0)$.
Let $p\in \mathbb N$ such that $\varphi(p)$ is defined, and $\varphi(p)\in \{n\in \mathbb N|u_p\leq a_n\leq v_p\}$
The set $\{n\in \mathbb N|u_{p+1}\leq a_n\leq v_{p+1}\}$ is infinite. Therefore, the set $$\{n\in \mathbb N| (n\geq\varphi(p) +1) \wedge (\ u_{p+1}\leq a_n\leq v_{p+1})\}$$ is also infinite. (Since its complement in the first set is always finite, and their union is infinite.)
We define $\varphi(p+1)$ as any element of that set. It satisfies the recurrence condition and $\varphi(p+1)>\varphi(p)$. Therefore we can completely define a strictly increasing function $\varphi$ which satisfies:
$$\forall p\in \mathbb N:\varphi(p)\in \{n\in \mathbb N|u_p\leq a_n\leq v_p\}$$
Therefore:
$$\forall p\in \mathbb N: u_p\leq a_{\varphi(p)}\leq v_p$$
And $(a_{\varphi(n)})_{n\in \mathbb N}$ is a convergent sequence.
Proof 2: that every real Cauchy sequence is convergent.
Let $(a_n)_{n\in \mathbb N}$ be a Cauchy sequence. We will first prove that it is bounded.
$\exists N\in \mathbb N: \forall n\geq N: |a_N-a_n|<1$
So, $\forall n\geq N: |a_n|<1+|a_N|$
We set $M=\max\{|a_0|,|a_1|,\dots, |a_{N-1}|, |a_N|+1\}$
Then $\forall n\geq N:|a_n|\leq M$
And $\forall n\leq N:|a_n|\leq M$
So $(a_n)_{n\in \mathbb N}$ is bounded.
Then it has a convergent subsequence. That is, there exists $\varphi:\mathbb N \rightarrow\mathbb N $ which is strictly increasing and $(a_{\varphi(n)})_{n\in \mathbb N}$ converges to $l$.
We will now prove that $(a_n)_{n\in \mathbb N}$ converges to $l$.
Let $\varepsilon>0$
$$\exists N_1\in \mathbb N: \forall n\geq N_1:|a_{\varphi(n)}-l|<\varepsilon/2$$ $$\exists N_2\in \mathbb N: \forall n,m\geq N_2:|a_n-a_m|<\varepsilon/2$$
We set $N= \max(N_1,N_2)$. Then, since $\forall n\in \mathbb N:\varphi(n)\geq n$:
$$\forall n\geq N: |a_n-l|\leq |a_n-a_{\varphi(n)}|+|a_{\varphi(n)}-l|<\varepsilon$$
Therefore, $(a_n)_{n\in \mathbb N}$ converges to $l$.
$\endgroup$ $\begingroup$A Cauchy sequence (on $\mathbb{R}$), $X =\{x_i\}$ has the property that $$\forall \epsilon\gt0, \exists N s.t. \forall n,m\ge N,|x_n-x_m|\lt\epsilon$$. If $X$ doesn't converge, then there must be $L_1$ and $L_2$ with $L_1= L_2-\delta$ such that $\delta\gt0$, and $X$ must achieve these limits infinitely often, which contradicts the definition of a Cauchy sequence.
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