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I know one proof of Bolzano's Theorem, which can be sketched as follows:


  • $f$ a continuous function in $[a,b]$ such that ${f(a)<0<f(b)}$.

  • ${A=\{x:a<x<b \text{ and } f <0\in[a,x] \}}$

    1. $A \neq \emptyset $
    2. $\exists \delta : a\leq x < a+\delta \Rightarrow x \in A $
    3. $ b$ is an upper bound and $\exists \delta :b-\delta <x \leq b $ and $x$ is another upper bound of $A$.

From the previous observations, $A$ has a supremum $\alpha$, from which we show $f(\alpha) =0$ ad absurdum.

Suppose $f(\alpha) <0$. Then

$$\exists \delta : \alpha - \delta <x<\alpha +\delta \Rightarrow f(x) <0 $$ Since $\alpha$ is the l.u.b., $$\exists x_0 : \alpha - \delta <x_0<\alpha$$ or else $\alpha$ wouldn't be the l.u.b.

Then $f<0$ in $[a,x_0]$. But if $\alpha < x_1 < \alpha +\delta$ then $f$ is also negative in $[x_0,x_1]$. Thus $f$ is negative in $[a,x_1]$, so $x_1 \in A$, which can't happen since $\alpha$ was the supremum.

The same procedure is used to rule out $f(\alpha) >0$, from where it is concluded that $f(\alpha) =0$.

My main concerns are:

  • Is the theorem necesserailly proven using the lub property of $\mathbb R$? (I suppose so).
  • How could another proof be constructed?
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Is lub-property a needed tag? This appears to be the only question using it. – Dylan Moreland May 12 '12 at 19:22
@DylanMoreland I thought it is an important property of $\Bbb R$. If it is not useful, just replace it by something similar. – Pedro Tamaroff May 12 '12 at 19:24
This may be of interest. – David Mitra May 12 '12 at 19:27
up vote 3 down vote accepted

You need the least upper bound property of $\mathbb R$, because since $\mathbb R$ is essentially $\mathbb Q$ closed under the operation of "taking l.u.b."'s, to see the necessity of it, you only need to notice that the result is not true over $\mathbb Q$ : take a continuous function $f : \mathbb R \to \mathbb R$ that has a unique zero $\alpha$ where $\alpha \notin \mathbb Q$ and $f(\alpha) = 0$, and then restrict your function to $\mathbb Q$, i.e. consider the same function but now $f : \mathbb Q \to \mathbb R$. Your function is still continuous over $\mathbb Q$, but it has no zeros anymore. Therefore, the l.u.b. property is necessary to do this proof.

I believe this answers your second question as well.

Hope that helps,

share|cite|improve this answer
For the second a simple sketch proof would suffice. I was quite sure I needed the lub properting. Never thought of $\Bbb R$ as closed under the "taking lub" operation, very enlightening. But then, wouldn't the operation be from a set $A \subset \Bbb R $ to $\Bbb R$ itself? As in $\star:A \mapsto \Bbb R$? – Pedro Tamaroff May 12 '12 at 19:33
The idea of closing under l.u.b.'s is this : consider all bounded above subsets of $\mathbb Q$ that do not contain their lubs in themselves, and for every such subset of $\mathbb Q$ that does not contain its lub, call this subset $E$, define the element $e$ such that $e$ is an upper bound of $E$ and $e$ is a lower bound of the set of all upper bounds of $E$ (this element is not in $e$ by construction, but will be in a new set that you will call $\mathbb R$). The fact that the sets of all such upper bounds gives you the set $\mathbb R$, is essentially Dedekind's construction. – Patrick Da Silva May 12 '12 at 22:40
You need to take care of the details though, this is just a rough statement. If you really wanted to "close under lub" like this, by essentially "adding elements until you're closed" like we usually close things in algebra, you need to quotient the new set you get by the equivalence relation "of those sets in $\mathbb Q$ that have the same lub". The point here is not to do this construction, but seeing that it is possible to do it enlightens us to find the necessity for the lub property of $\mathbb R$ in your situation. – Patrick Da Silva May 12 '12 at 22:44
Thank you very much Patrick! This really helped! – Pedro Tamaroff May 13 '12 at 17:17

You really do need the completeness of $\Bbb R$ (the lub property). If you remove just one point $c$ from $\Bbb R$, you can easily find a counterexample to the intermediate value theorem: just define $$f:[c-1,c+1]\to\Bbb R:x\mapsto\begin{cases}-1,&\text{if }c-1\le x<c\\1,&\text{if }c<x\le c+1\;.\end{cases}$$ The problem here, of course, is that the set $(\leftarrow,c)$ has no supremum in $\Bbb R\setminus\{c\}$.

The same phenomenon can be duplicated for any linear order. Let $\langle X,\le\rangle$ be a linear order. To keep the description simple, I'll assume that $X$ has no endpoints, but this is not essential. Let $\tau$ be the order topology on $X$, the topology whose open sets are $\varnothing$ and arbitrary unions of open intervals in $X$.

Suppose that $A\subseteq X$ is bounded above but has no supremum. Fix $a\in A$ and an upper bound $b$ of $A$; clearly $a<b$. Let $L=\{x\in[a,b]:\exists a'\in A\,(x\le a')\}$; then $a\in L$, $b\notin L$, and for each $x\in L$, $[a,x]\subseteq L$. It's easy to see that if $L$ had a supremum $s$, $s$ would also be the supremum of $A$. Thus, $L$ has no supremum, and therefore $$L=\bigcup_{x\in L}[a,x)\quad\text{ and }\quad[a,b]\setminus L=\bigcup_{x\in[a,b]\setminus L}(x,b]$$ are open subsets of $[a,b]$.

Now define $$f:[a,b]\to\Bbb R:x\mapsto\begin{cases}-1,&\text{if }x\in L\\1,&\text{if }x\in[a,b]\setminus L\;.\end{cases}$$

Because both $L$ and $[a,b]\setminus L$ are open in $[a,b]$, $f$ is continuous, but it jumps from $-1$ to $1$ without passing through $0$.

share|cite|improve this answer
Very good generalization of the argument. +1! – Patrick Da Silva May 12 '12 at 22:45
What's a linear order? And the order topology $\tau$? – Pedro Tamaroff May 18 '12 at 1:22
@Peter: $\langle X,\le\rangle$ is a linear order, also called a total order, if $\le$ is reflexive, transitive and total: for every $x,y\in X$, $x\le y$ or $y\le x$. There’s a brief introduction to the order topology here that should at least get you started. – Brian M. Scott May 18 '12 at 5:33

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