Can the Intermediate Value Theorem be proved by Heine-Borel Lemma? Can the Intermediate Value Theorem be proved by Heine-Borel Lemma, and how?
I mean "Every open cover of close interval has a finite subcover", without compactness etc. 
Because in class we proved it via Cantor's Lemma... But Cantor's Lemma gives me a certain point in the function domain that I can work with, but how do I find this point with Hiene-Borel? 
Thanks
 A: In reverse mathematics, the intermediate value theorem is true in a weaker axiom system, $RCA_0$, than Heine-Borel, which requires $WKL_0$.
This means that there are models of weak second-order theories of numbers in which IVT is true but Heine-Borel is not.
That doesn't mean you can't prove it from Heine-Borel, just that the IVT is more"basic."
(Indeed, $WKL_0$ is equivalent to $RCA_0$ with Heine-Borel added, and it is known to be strictly stronger axiom system - the Weak König's Lemma is provably independent from $RCA_0$.)
A: My first thought is no, because the Intermediate Value theorem and Heine-Borel theorem leverage different properties of the unit interval, connectedness and compactness, respectively. But let's see if we can turn the IVT into something using compactness:
Let $f:[a,b]\to\Bbb R$ be a continuous function, and let $f(a)<c<f(b)$. Define $g(x)=|f(x)-c|$; then $g$ is also continuous, and by the extreme value theorem (which can be proven using Heine-Borel) $g$ takes a minimum at some $y$. If $g(y)=0$ we are done, otherwise $g(y)=r>0$ and the range of $f$ misses the interval $(c-r,c+r)$.
Well, we managed to widen the "hole" in the range of $f$ from a point to an interval, but we still haven't gotten any closer to the contradiction. Of course we could consider the smallest $x$ such that $f(x)\ge c$, but then we would be back to the standard proof. Interestingly, the point $y$ is actually a root; it just doesn't have all the properties that we need to finish the proof (being a minimal root does the trick).
A: Heine-Borel is equivalent to compactness for $\Bbb R^n$ spaces. Here is a proof using compactness.
Theorem. Let $f:[a,b]\to \Bbb R$ be a continuous function. If $f(a)f(b) \lt 0$, then there exists a point $c$ in $[a,b]$ such that $f(c)=0$.
Proof. Without loss of generality, assume $f(a)\lt 0\lt f(b)$. Suppose that for each point $x\in[a,b]$, we have $f(x)\neq 0$. By continuity, for each $x\in[a,b]$ there's a $\delta_x\gt 0$ such that $f$ does not change sign in $I_x = (x-\delta_x,x+\delta_x)$. We can take $\delta_x$ so small that at least one of the endpoints of $I_x$ lies within $[0,1]$.
Notice that if $f(x)\gt 0$ and $f(y)\lt 0$ then $I_x\cap I_y = \emptyset$
So the collection $\{I_x\}_{x\in[a,b]}$ is an open cover for $[a,b]$ which is compact, so there's a finite subcollection say $\{I_j\}_{j=1}^n$ which cover $[a,b]$. Let's name them from left to right, namely
\begin{gather*}
I_j = (a_j,b_j) \\
a_1\lt\cdots\lt a_n\\
b_1\lt\cdots\lt b_n
\end{gather*}
in particular $a\in I_1$ and $b\in I_n$.
Now, we divide the set of indexes in this way:
\begin{gather*}
N = \{j:x\in I_j\cap [a,b] \implies f(x)\lt 0\} \\
P = \{j:x\in I_j\cap [a,b] \implies f(x)\gt 0\}.
\end{gather*}
Take $m\in\{1,\dots,n\}$ such that $b_m = \max\{b_j : j\in N\}$. Then $1\leq m\lt n$ because $b\in I_n$ and $f(b)\gt 0$. Then $m+1\in P$ and so $I_m\cap I_{m+1}=\emptyset$. This implies that the interval $(b_m,a_{m+1})$ is not empty and clearly $$(b_m,a_{m+1})\not\subset\bigcup_{j=1}^n I_j.$$
By the construction the midpoints of $I_m$ and $I_{m+1}$ belong to $[a,b]$ the interval whose endpoints are the midpoints of $I_m$ and $I_{m+1}$ is contained in $[a,b]$ and this interval contains $(b_m,a_{m+1})$,so $(b_m,a_{m+1})\subset [a,b]\subset\bigcup_{j=1}^n I_j$. Contradiction.
