Is there a short proof for the Intermediate Value Theorem My final for my introductory analysis course is tomorrow and my teacher gave us a list of possible theorems to prove. If anyone could please show me a proof for The Intermediate Value Theorem that is short and easy to follow, so even if I still cannot understand it I can at least memorize it. Also, I have looked through numerous texts and the internet, but they all seem to confuse me. I know that itis an insult to all you math experts to memorize proofs, but I am desperate at this point. Thank you
 A: The important part is to apply the definition of continuity:
$f(x)$ is continuous at point $c$ if $\forall \epsilon>0, \exists \delta>0 \quad S.T \quad |x-c|<\delta \Rightarrow |f(x)-f(c)|<\epsilon$ (Alternatively speaking, there always exist $\delta>0$ such that for any $\epsilon>0,|f(x)-f(c)|<\epsilon $).
Proof:
Define $c=\sup\{x:f(x)\leq u\}$ ($\star$), and claim $f(c)=u$
Assume $x\in (-\delta+c,c+\delta),\delta>0$ ($x$ is in the $\epsilon$ neighborhood of $x$)
By definition of continuity $\forall \epsilon>0, |f(x)-f(c)|<\epsilon \Rightarrow -\epsilon+f(c)<f(x)<f(c)+\epsilon$
In the following, we will manipulate both side of this inequality and prove the intermediate value theorem by contradiction:

*

*If $f(c)>u$, then $f(c)-u>0$, so set $\epsilon=f(c)-u \Rightarrow f(x)>f(c)-\epsilon=f(c)-(f(c)-u)=u$ (use the left side of the inequality) $\Rightarrow \forall x\in (-\delta+c,c+\delta),f(x)>u$, so it means that $(c-\delta)$ is the least upper bound of the set $\{x:f(x)\leq u\}$, which contradicts with the definition of $c$(also a least upper bound and it's not possible to have 2 least upper bounds at the same time.)

*If $f(c)<u$, then $u-f(c)>0$,so set $\epsilon=u-f(c) \Rightarrow f(x)<f(c)+\epsilon=f(c)+(u-f(c))=u$ (use the right side of the inequality) $\Rightarrow \forall x\in (-\delta+c,c+\delta),f(x)<u$. This means that there exist $x>c$ such that $f(x)<u$, which contradicts with the definition of $c$ again. (because $c$ is the sup of the set $\{x:f(x)\leq u\}$)

If you set $u=0$, then it's the answer of your question. Hope this can help.
A: This theorem is a case when the most intuitive proof requires relatively advanced technique. here's a proof using general topology.
Consider a continuous function $f: \mathbb{R} \to \mathbb{R}$ which takes values in $a, b$ but not $c \in (a, b)$. Then $f$ factors through the embedding $i: \mathbb{R} \setminus \{c\} \to \mathbb{R}$: $f = i \circ \hat{f}$, where $\hat{f}: \mathbb{R} \to \mathbb{R} \setminus \{c\}$. $\hat{f}$ is continuous: any open set of $\mathbb{R} \setminus \{c\}$ is obtained by removing $c$ from an open subset of the whole $\mathbb{R}$. But since $f$ is nowhere equal $c$, removing this point doesn't affect the preimage, which is open. $\mathbb{R}$ is path-connected (and thus connected), but $\mathbb{R} \setminus \{c\}$ has two connected components. Therefore, the image of $\hat{f}$ must lie to the one side of $c$ - contradiction.
A: The indermediate value theorem says:

Let $f:[a,b]\to \mathbb{R}$ be continuous and $f(a)<0$ and $f(b)>0$, then there exists a 
  $\xi \in (a,b)$ such that $f(\xi)=0$.

You can prove it by using nested intervals:
You look at $f(\frac{a+b}{2})$, when it is bigger than null you look at $f$ on the interval
$[a,\frac{a+b}{2}]$, if it is smaller than 0 we look at $[\frac{a+b}{2},b]$, when it is $0$ we are done. Lets denote the left endpoints with $a_n$ and the right endpoints with $b_n$.
As the diameter of our nested intervals is $(b-a)\cdot 2^{-n}$ which clearly converges to zero we have 
$$\lim_{n \to \infty} a_n =\lim_{n\to \infty} b_n=\xi$$
As $f$ is continuous we get 
$$\lim_{n\to \infty} f(a_n)=\lim_{n\to \infty} f(b_n)=f(\xi)$$
On the other hand we know 
$$f(a_n) < 0 \quad \forall n$$
and $$ f(b_n)>0 \quad \forall n $$
Hence we know 
$$\lim_{n\to \infty} f(a_n)\leq 0$$
and $$\lim_{n\to \infty} f(b_n)\geq 0$$
Hence
$$0\leq f(\xi) \leq 0$$ 
Hence $f(\xi)=0$ 
You use that when $C_i$ is closed, bounded  and non empty for all $i$ and $C_{i+1} \subset C_i$ for all $i$ then 
$$\bigcap_{i \in \mathbb{N}} C_i \neq \varnothing$$
A: Let $f : [a,b]\to \mathbb R$ be continuous, $f(a)< 0, f(b) > 0$. Because $[a,b]$ is connected, and $f$ continuous, $f([a,b]) \subset \mathbb R$ is connected. But the only connected sets in $\mathbb R$ are intervals, so $I=f([a,b])$ is an interval with $I \ni f(a) < 0$ and $I \ni f(b) > 0$. Thus $0\in I = f([a,b])$, so there exists a $\xi \in [a,b]$, $f(\xi ) = 0$.
A: Let $ f : [a, b] \longrightarrow \mathbb{R} $ be continuous, with $ f(a) < f(b) $. And say $ \eta $ is a real number such that $ f(a) < \eta < f(b) $.
We should now exhibit an $ x \in [a, b] $ at which $ f(x) = \eta $.
Let's call a point $ p $ in $ [a, b] $ "low" if $ f(p) < \eta $, and "high" if $ f(p) > \eta $. Let $ L $ be the set of all low points.
$ L $ contains $ a $ and is bounded above by $ b $, so $ l := \sup(L) $ exists.
Note $ a < l < b $ (Since $ a $ is low, by continuity at $ a $ there is a $ \delta > 0 $ such that all points of $ [a, a + \delta] $ are low. So $ a < a + \delta \leq l $, giving $ a < l $. Similarly $ l < b $ too).
Is $ l $ low ? If it were, by continuity at $ l $ there would be a $ \delta > 0 $ such that all points of $ [l - \delta, l + \delta] $ are low. But by definition of $ l $, no point $ > l $ can be low, a contradiction.
Is $ l $ high ? If it were, by continuity at $ l $ there would be a $ \delta > 0 $ such that all points of $ [l-\delta, l+\delta] $ are high. But by definition of $ l $, there must be a point in $ [l-\delta, l] $ which is low, a contradiction.
So $ l $ is neither low nor high, i.e. $ f(l) = \eta $, as needed.
A: Here is a short proof via Heine Borel Theorem. We need to establish the following:
Intermediate Value Theorem: Let $f$ be continuous on $[a, b]$ and let $f(a)f(b) < 0$. Then there is a point $c \in (a, b)$ for which $f(c) = 0$.
Let's extend definition of $f$ beyond $[a, b]$ by setting $f(x) = f(a)$ for $x < a$ and $f(x) = f(b)$ for $x > b$. Then $f$ is continuous everywhere. Let's now assume on the contrary that $f$ does not vanish anywhere in $(a, b)$.
By continuity of $f$ on each point $x \in [a, b]$ there is a neighborhood $I_{x}$ of $x$ on which $f$ maintains a constant sign. The collection of such intervals $I_{x}$ forms an open cover for $[a, b]$ and hence by Heine-Borel Theorem a finite number of such intervals cover $[a, b]$.
The end points of these chosen finite intervals $I_{x}$ which lie in $[a, b]$ form a partition of $[a, b]$ say $$P = \{a = x_{0}, x_{1}, x_{2}, \ldots, x_{n} = b\}$$ and let $\delta = \max_{i = 1}^{n}(x_{i} - x_{i - 1})$ be the norm of partition $P$.
If we have another partition $P' = \{x_{0}', x_{1}', \ldots, x_{m}'\}$ of $[a, b]$ such that norm of $P'$ is less than $\delta$ then each sub-interval $[x_{j - 1}', x_{j}']$ generated by $P'$ lies completely inside one of the chosen intervals $I_{x}$. Hence $f$ maintains constant sign on each of the sub-intervals $[x_{j - 1}', x_{j}']$ and therefore is of constant sign on $[a, b]$. This contradiction shows that $f$ must vanish at some point $c \in (a, b)$.
