I'm trying to do a series of exercises from Spivak's Calculus, in chapter 8, Least Upper Bounds.
I'm trying to tackle these two exercises, $5.$ and $^*.6$
From $5.$ I have proven the first claim
$(a)$ Let $x-y>1$. Prove there is an integer $k$ such that $x<k<y$.
P Let $\ell$ be the greatest integer such that $\ell \leq x$. Then
$$y-x >1$$
$$y-\ell >1$$
$$y>1+\ell $$
Thus the integer $\ell +1$ is between $y$ and $x \text{ }\blacktriangle$.
$(b)$ Let $x<y$. Show there is a rational $r$ such that $x<r<y$.
P If $x<y$ then there is $\epsilon>0$ such that $x+\epsilon=y$. Thus $\epsilon=y-x$. But from T3, we have that there is an $n$ such that $1/n < \epsilon$, thus
$$\frac{1}{n} < y-x$$
$$1<ny-nx \text{ }\blacktriangle$$
and from the last theorem we have that there is a integer $k$ such that
$$nx<k<ny$$
$$x< \frac{k}{n}<y \text{ }\blacktriangle$$
$(c)$ Let $r<s$ be rational numbers. Prove there is an irrational number between $r$ and $s$. Hint: it is known there is an irrational number between $0$ and $1$.
Ok, this is a proof based on your answers.
P Since $\sqrt 3 $ is irrational and $\sqrt{3}<3$, then $\ell = \sqrt{3}/3<1$ and thus it is in $[0,1]$.
Now, $r<s \Rightarrow 0<s-r$. Then
$$0<\ell < 1$$
$$0<\ell(s-r) < s-r$$
$$r<r+\ell(s-r) < s$$
And since $\ell$ is irrational $r+\ell(s-r)$ is irrational. $\blacktriangle $
$(d)$ Show that if $x<y$ then there is an irrational number between $x$ and $y$: There is no need to work here, this is consequence of $(b)$ and $(c)$
This one was quite straightforward, but thanks anyways.
P
$$x<y$$
$$(b)\Rightarrow x<r<y,r<y \Rightarrow r<q<y \Rightarrow x<r<q<y$$
Then by $(c)$, there is an irrational $\ell$ such that
$$ x<r<\ell<q<y \text{ } \blacktriangle $$
This will let me conclude
$\mathbb Q$ is dense on any $[a,b]\subset \Bbb R$
$\mathbb I$ is dense on any $[a,b] \subset \Bbb R $
and will let me move on into $^*6.$ which is
$(a)$ Show that $f$ is continuous and $f(x)=0$ for all $x$ in a dense set $A$, then $f$ is $f(x)=0$ for all $x$.
$(b)$ Show that $f$ and $g$ are continuous and $f(x)=g(x)$ for all $x$ in a dense set $A$, then $f(x)=g(x)$ for all $x$.
$(c)$ If we suppose $f(x)\geq g(x)$ for all $x$ in $A$, then $f(x)\geq g(x)$ for all $x$. ¿Can $\geq$ be substituted with $>$ everywhere?
I'm not asking for solutions for this last problems (which will be asked separately), but for $(c)$ and $(d)$ in $5.$
The chapter has several important proofs, which might or might not be relevant here, but I think it is important you know what tools we have at hand:
THEOREM 7-1 If $f$ is continuous on $[a,b]$ and $f(a)<0<f(b)$, then there is $x \in [a,b]:f(x)=0$
THEOREM 1 If $f$ is continuous in $a$, then there exists a $\delta>0$ such that $f$ is bounded above in $(a-\delta,a+\delta)$.
THEOREM 7-2 If $f$ is continuous on $[a,b]$ then $f$ is bounded on $[a,b]$.
THEOREM 7-3 If $f$ is continuous on $[a,b]$ then there is an $y$ in $[a,b]$ such that $f(y)\geq f(x)$ for all $x$ in $[a,b]$.
THEOREM 2 $\Bbb N$ is not bounded above.
THEOREM 3 If $\epsilon >0$, there is an $n \in \Bbb N$ such that $1/n < \epsilon$.
