I've learned that the Heine-Borel theorem and the Bolzano-Weierstrass theorem are both the equivalent form of the completeness axiom, but I have difficult in proving this.
Specifically, my book gives me the hint to prove Archimedes principle first, and I proceed as follows:
For Heine-Borel deduce Archimedes:
Assume $\mathbb{N}$ is bounded. Then there exist $m\in \mathbb{R}$ so that $\forall n\in \mathbb{N}$, $n<m$.
Consider a closed interval $[a,b]$ and open intervals $I_{\alpha}(x-\frac{b-a}{3m}, x+\frac{b-a}{3m})$, $x\in [a,b]$. Then from Heine-Borel theorem there exist open intervals $I_1,I_2, \dots ,I_n$ that cover $[a,b]$, where $n\in \mathbb{N}$.
Then $n>m$, otherwise the total length of those intervals$\le \frac{2m(b-a)}{3m}< b-a$, contradiction.
So there exists a natural number $n$ greater than $m$, contradiction.
And for the Bolzano-Weierstrass to prove Archimedes I did:
If natural numbers have upper bound then it is a bounded infinite set. Then from Bolzano-Weierstrass there's some point it converges. Assume this point to be $n\in \mathbb{N}$. But from the natural number's property it gives that any two natural numbers differ by at least 1; that is, $(n-\alpha, n+\alpha)$ contains no natural numbers when $\alpha <1$, contradiction.
Are my proof correct? If they were wrong, could you point out the mistake and give me a correct one? Thanks in advance.
You can use $[0,1]$, instead of a generic interval $[a,b]$. The set of intervals of the form $$ I_x=\left(x-\frac{1}{m},x+\frac{1}{m}\right) $$ obviously covers $[0,1]$. If $I_{x_1}\cup\dots\cup I_{x_n}$ covers $[0,1]$, then $$ 1\le \frac{2n}{m} $$ a contradiction.
For the second part, the idea is good, but you should say that the infinite bounded set $\mathbb{N}$ has a limit point (or accumulation point, depending on terminology). If this point is $x$, then in each of its neighborhoods, for instance $(x-1/2,x+1/2)$, there should be infinitely many natural numbers, a contradiction, because it can contain at most one. Note that you're not allowed to say that $x\in\mathbb{N}$.