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I need to find examples of the following, but I feel like they're trick questions.

  1. A unbounded function on a bounded closed interval (the function must be defined at every point in the interval)

I was thinking something like 1/x, but it's not defined at 0 (it'd need to be (0,1] ), so I don't see how I can provide something that is defined at the boundary points and unbounded on the interval. I would think I'd need something that blows up, but then it won't be defined at every point in the interval

  1. f : [0, 1] → [0, 1], having the intermediate-value property but continuous at only one point

For a function to experience the intermediate-value property, doesn't it need to be continuous on its domain (the whole closed interval)? How can I provide an example when it's only continuous at one point?

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  • $\begingroup$ There is no continuous function $f$ on a closed, bounded interval $I$ satisfying (1). Since $I$ is compact, so is $f(I)$, and in particular $f$ is bounded on $I$. $\endgroup$ – Travis Oct 18 '15 at 19:11
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    $\begingroup$ But, in 1., it is not demanded that $f$ be continuous. $\endgroup$ – Daniel Fischer Oct 18 '15 at 19:11
  • $\begingroup$ But, it has to be defined at all points, so how could I make 1. blow up? $\endgroup$ – user269711 Oct 18 '15 at 19:14
  • $\begingroup$ what about 0 if x=0 and 1/x if x!=0 on the interval [0,1]? $\endgroup$ – user269711 Oct 18 '15 at 19:14
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    $\begingroup$ Your title is misleading. You may want to change it. $\endgroup$ – Aloizio Macedo Oct 18 '15 at 19:22
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f(x) = 0 if x = 0. f(x) = 1/x if x $ne$ 1.

Unbounded, defined everywhere, but not continuous.

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    $\begingroup$ Another I like but simply not as straightforward is f (m/n) = n for rationals. It's Unbounded everywhere but it's tedious to show. $\endgroup$ – fleablood Oct 18 '15 at 22:38
  • $\begingroup$ Ah, that makes sense! Is it possible to think of an example for 2)? I've been struggling to think of one for that for awhile. $\endgroup$ – user269711 Oct 18 '15 at 22:50
  • $\begingroup$ Hermes answer for 2) is good. Let f (x) = x if x is rational. Let f (x) = 1-x if x is not rational. If f (a)< c < f (b) then there is an x that f (x) = c, it just isn't that x is between a and b. $\endgroup$ – fleablood Oct 18 '15 at 22:56
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    $\begingroup$ @fleablood $x$ must be between $[a,b]$, that's the whole point with IVP. A discontinuous function to satisfy IVP must be really nasty. $\endgroup$ – A.Γ. Oct 18 '15 at 23:05
  • $\begingroup$ AG. I agree that is version of IVT has no "point" but that wasn't a requirement of the problem. This satisfies the letter but not the spirit of the IVT. I think if we require that x be between a and b, then f must be continuous but I'm not willing to figure that out using a phone rather than pencil and paper. $\endgroup$ – fleablood Oct 18 '15 at 23:26
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The second example:

  1. Take the Conway base 13 function $F(x)$, which is known to take on every real value on each interval (and is, therefore, nowhere continuous), and the interval $(0,1)$.
  2. Consider $g(t)=\frac12+\frac{1}{\pi}\arctan t$. The function $g\circ F$ takes on every value in $(0,1)$ on each real interval in $(0,1)$.
  3. Define $$ f(x)=\left\{ \begin{array}{ll} 0, &\text{ if } x=0,\\ 1, &\text{ if } x=1,\\ x\cdot (g\circ F)(x)&\text{ if } x\in(0,1). \end{array} \right. $$ The function $f\colon[0,1]\to[0,1]$ takes on every real value in $(0,b)$ on each interval $(a,b)\subset[0,1]$ and, thus, satisfies IVP on $[0,1]$ and continuous only at $x=0$.
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  • $\begingroup$ so the hard part is just to understand how the Conway function works, your answer is much nicer than the mentioned function! Maybe you could add to "1." that the function is indeed nowhere continuous. $\endgroup$ – user190080 Oct 19 '15 at 0:05
  • $\begingroup$ @user190080 Thank you, I have added the line about being nowhere continuous. $\endgroup$ – A.Γ. Oct 19 '15 at 0:19
  • $\begingroup$ I'm pretty sure that the example works, but how do you justify that $f$ takes on every real value in $(0,b)$? (Silly analogy: $1/x$ takes every non-zero real value, but $x\cdot 1/x$ does not.) $\endgroup$ – mrf Oct 19 '15 at 11:09
  • $\begingroup$ @mrf Right. To approach zero is not a question ( $b\cdot 0\to 0$). To get arbitrary close to $b$: take any $c\colon a<c<b$ then on the interval $(c,b)$ the scaling of $(0,1)$ by $x$ will give you larger values than $c$. For example, on $((c+b)/2,b)$ you will get infinitely many values in $(0,(c+b)/2)\ni c$. $\endgroup$ – A.Γ. Oct 19 '15 at 12:04
  • $\begingroup$ @mrf The point here is that the Conway function covers all values in $\mathbb{R}$ on any interval kind of uniformly, i.e. independently on the interval bounds. $\endgroup$ – A.Γ. Oct 19 '15 at 12:23
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I will attempt the second part. To do this I will assume the result that the Cantor set can be mapped surjectively onto $[0,1]$ (which can be done, for example, by sending ternary expansions with 0's and 2's into binary expansions with the corresponding 0's and 1's).

We will construct a set $S \subset [0,1]$ which can be written as the union of countably many disjoint Cantor sets $C_{n,k}$, where each $C_{n,k}$ is contained in the dyadic interval $[k2^{-n}, (k+1)2^{-n}]$. The term "Cantor set" refers to a translate and dilate of the usual $1/3$-Cantor set in $[0,1]$.

If we can construct this set $S$ then we would be done. Indeed, for each $n,k$ there exists a surjection $\phi_{n,k}:C_{n,k} \to [0,1]$. Then we can define a map $\phi: [0,1] \to [0,1]$ as $$\phi(x)=\begin{cases} 0, \;\;\;\;\;\;\;\;\text{ if } x \notin S \\ \phi_{n,k}(x), \;\text{ if } x \in C_{n,k} \text{ for some } n,k \end{cases}$$ This function is discontinuous at every point and has IVT, since any subinterval of $[0,1]$ must contain one of the $C_{n,k}$, which is mapped onto $[0,1]$ by $\phi$. If you want continuity at exactly $1$ point, then just consider $x\phi(x)$.

To finish, we will now construct the sets $C_{n,k}$, for $n \geq 0$ and $k\in \{0,...,2^n-1\}$. We apply induction on $n$ for this construction. First, let $C_{0,0}$ be the usual Cantor set in $[0,1]$. For the inductive step of the construction, suppose that $N \geq 1$ and that we have constructed disjoint Cantor sets $C_{n,k} \subset [k2^{-n},(k+1)2^{-n}]$ for all $k \in \{0,...,2^n-1\}$ with $n<N$. Since each of the sets $C_{n,k}$ is closed and nowhere dense, it follows that the union of these $C_{n,k}$ (for $n<N$ and $0\leq k \leq 2^n-1$) is also closed and nowhere dense. Hence each dyadic interval $[k2^{-N},(k+1)2^{-N}]$ contains an open interval $J_{N,k}$ not intersecting any of these $C_{n,k}$. Now for each $0 \leq k \leq 2^N-1$, just define $C_{N,k}$ to be any Cantor set contained in $J_{N,k}$. Clearly these $C_{N,k}$ are disjoint from the previously constructed $C_{n,k}$. Thus we have defined disjoint Cantor sets $C_{n,k}\subset [k2^{-n},(k+1)2^{-n}]$ for all $0 \leq k \leq 2^n-1$ with $n< N+1$. So continue inductively...

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