I'm reading my multivariable calculus lecture notes and have some questions$\dots$
Here are some Theorems mentioned in the notes: $$\boxed{\begin{align}\text{MCT (Monotone Convergence Theorem)}\\ \text{BST (Bounded Squence Theorem)}\end{align}}$$
Consider the case in $\mathbb{R}$ we have:
$\text{Every bounded sequence in $\mathbb{R}$ has a subsequence that converges to a limit.}\tag*{BST}$
$\text{Every non-decreasing sequence of real numbers}\tag*{MCT}\\ \text{that is bounded above converges to a limit.}$
$\text{Any non-empty set of real numbers} \\\text{that has an upper bound must }\\\text{have a least upper bound in real numbers.}\tag*{Dedekind Completeness}$
You can check that these are all actually equivalences, in other words, that you can prove the completeness of $\mathbb{R}$ (the least upper bound property) from the MCT and that you can prove the MCT from the BST:
$$\text{BST}\Leftrightarrow \text{MCT}\Leftrightarrow\text{Completeness}$$
Hence in $\mathbb{R}$ the notion of completeness is equivalent to the Bounded Sequence Theorem. Finally notice that the statement of the Bounded Sequence Theorem no longer requires the definition of a least upper bound (or an order between vectors) and has a generalization to the Bounded Sequence Theorem in $\mathbb{R}^n$ as above.
This allows us to define completeness of $\mathbb{R}^n$ in terms of the BST. The analogue of the completeness axiom in higher dimensions thus becomes:
$\mathbb{R}^n$ is complete if Every bounded sequence in $\mathbb{R}^n$ has a convergent subsequence.
$\dots$ We could also show that completeness is equivalent to the $\underline{\text{Intermediate Value Theorem}}$, or to the statement that $\underline{\text{Every absolutely convergent sequence converges}}$.
I think it's saying since for $\mathbb{R}$ we have Dedekind Completeness (the least upper bound property) , which is hard to extend to $\mathbb{R}^n$, but BST $\Leftrightarrow$ Completeness, so we can take BST as an alternative definition for Completeness.
I believe $\text{BST}\Leftrightarrow \text{MCT}\Leftrightarrow\text{Completeness}$ actually means:
$$(\text{BST}\Leftrightarrow \text{MCT})\land(\text{BST}\Leftrightarrow\text{Completeness})\land(\text{MCT}\Leftrightarrow\text{Completeness})$$
And the note also listed the proof outline for $\mathbb{R}$ which might be: (Any cycle would work)
$$(\text{BST}\Rightarrow\text{MCT})\land (\text{MCT}\Rightarrow\text{Completeness})\land(\text{Completeness}\Rightarrow\text{BST})$$
I'm not sure why it's not taking Cauchy Completeness for $\mathbb{R}^n$, since $$\text{Cauchy Completeness}\Leftrightarrow\text{Dedekind Completeness}$$
Is MCT also an alternative definition for Completeness $?$
How does $\text{BST}\Leftrightarrow \text{MCT}\Leftrightarrow\text{Completeness}$ generalise to $\mathbb{R}^n?$
My attempts:
$\text{Every bounded sequence in $\mathbb{R}^n$ has a subsequence that converges to a limit.}\tag*{BST}$
$\text{Every non-decreasing sequence in $\mathbb{R}^n$}\tag*{MCT}\\ \text{that is bounded above converges to a limit.}$
$\text{Every Cauchy sequence in $\mathbb{R}^n$ converges}\tag*{Cauchy Completeness}$
In order to prove $\text{BST}\Leftrightarrow \text{MCT}\Leftrightarrow\text{Completeness}$, maybe take the following proof outline:
$$(\text{BST}\Rightarrow\text{Completeness})\land (\text{Completeness}\Rightarrow\text{MCT})\land(\text{MCT}\Rightarrow\text{BST})$$
For BST $\Rightarrow$ Completeness:
First prove two lemmas
$1.$ Every Cauchy sequence are bounded in $\mathbb{R}^n$
$2.$ Every Cauchy sequence in $\mathbb{R}^n$ has a convergent subsequce are convergent.
Then we take arbitrary Cauchy sequence, apply lemma $1$ that it's bounded, then apply BST so it has a subsequence that converges to a limit, finally apply lemma $2$ have it converges, hence Completeness hold.
Some details for each Lemma:
Lemma $1$: Cauchy sequence in $\mathbb{R}^n$ is bounded
Proof.
Assume $\{a_j\}_{j=1}^\infty$ is a Cauchy sequence
$$\forall\varepsilon>0,\exists N\in\mathbb{N},s.t.\forall j,k\in\mathbb{N},(j,k\ge N\rightarrow\Vert a_j-a_k\Vert <\varepsilon)$$
Show it’s bounded
$$\exists r>0,s.t.\forall j\in\mathbb{N},\Vert a_j\Vert <r$$
Let $S:=\{d\in\mathbb{R}:\exists i\in [1,k],s.t.\Vert 0-a_i\Vert= d\}$
Also $M:=\max(S)$
And $r:=M+\varepsilon$
$\underline{\text{Case} 1:j\le k}$
Then we have
$$j\in[1,k]\Rightarrow \Vert 0-a_j\Vert \le M$$
$$\Rightarrow \Vert a_j\Vert =\Vert 0-a_j\Vert < M+\varepsilon=r$$
$$\Rightarrow \boxed{\Vert a_j\Vert <r}$$
$\underline{\text{Case} 2:j>k}$
Since $k\in[1,k]$ we have
$$a_k\le M\Rightarrow\Vert 0-a_k\Vert \le M$$
And by assumption
$$\Vert aj-a_k\Vert <\varepsilon$$
Together with Triangle inequality of norm in $\mathbb{R}^n$ implies
$$\Vert a_j\Vert =\Vert 0-a_j\Vert \le\Vert 0-a_k\Vert +\Vert a_k-a_j\Vert <M+\varepsilon=r$$
$$\Rightarrow\boxed{\Vert a_j\Vert <r} \tag*{$\square$}$$
Lemma $2$: If Any Cauchy sequence in $\mathbb{R}^n$ has a convergent subsequce then that Cauchy sequence is convergent.
Proof.
Assume $\{a_j\}_{j=1}^\infty$ is a Cauchy sequence in $\mathbb{R}^n$ that has a convergent subsequence
$$\forall\varepsilon>0,\exists N\in\mathbb{N},s.t.\forall j,k\in\mathbb{N},(j,k\ge N\rightarrow\Vert a_j-a_k\Vert <\varepsilon)$$
$$\wedge\forall\varepsilon>0,\exists J> 0,s.t.\forall i\in\mathbb{N}(i\ge J\rightarrow\Vert a_{j_i}−L\Vert <\varepsilon)$$
Show $\{a_j\}_{j=1}^\infty$ converges to the same point.
$$\forall\varepsilon>0,\exists K> 0,s.t.\forall j\in\mathbb{N}(j\ge K\rightarrow\Vert a_{j}−L\Vert <\varepsilon)$$
By rearrange the assumption we have
$$\forall\frac{\varepsilon}{2}>0,\exists N\in\mathbb{N},s.t.\forall j,i\in\mathbb{N},(j,i\ge N\rightarrow\Vert a_j-a_{j_i}\Vert < \frac{\varepsilon}{2})$$
$$\wedge\forall\frac{\varepsilon}{2}>0,\exists J> 0,s.t.\forall i\in\mathbb{N}(i\ge J\rightarrow\Vert a_{j_i}−L\Vert <\frac{\varepsilon}{2})$$
Let $K=\max\{N,J\}$, so we can use both assumptions
By Triangle inequality of norm in $\mathbb{R}^n$ implies
$$\Vert a_j-L\Vert =\Vert a_j-a_{j_i}+a_{j_i}-L\Vert \le\Vert a_j-a_{j_i}\Vert +\Vert a_{j_i}-L\Vert <\varepsilon$$
$$\Rightarrow \boxed{\Vert a_j-L\Vert <\varepsilon}\tag*{$\square$}$$
Is the idea correct $?$
Thanks for your help.