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
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
P If $x
$$\frac{1}{n} < y-x$$
$$1 and from the last theorem we have that there is a integer $k$ such that $$nx $$x< \frac{k}{n} $(c)$ Let $r Ok, this is a proof based on your answers. P Since $\sqrt 3 $ is irrational and $\sqrt{3}<3$, then $\ell = \sqrt{3}/3<1$ and thus it is in $[0,1]$. Now, $r $$0<\ell < 1$$ $$0<\ell(s-r) < s-r$$ $$r And since $\ell$ is irrational $r+\ell(s-r)$ is irrational. $\blacktriangle $ $(d)$ Show that if $x This one was quite straightforward, but thanks anyways. P $$x $$(b)\Rightarrow x Then by $(c)$, there is an irrational $\ell$ such that $$ x 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 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 $1/n < \epsilon$.