I am trying to do the following exercice :

1) Show that $f:\mathbb{R}\rightarrow[0,+∞)$,bijective, has an infinite number of discontinuity.

2) An example ?

My work:

1) I have succeeded by contradiction and use the fact that f continuous and injective is Strictly monotone
2) I have no idea..I can see it on a drawing but I was not able to explain that kind of function. If someone has an idea please share it.

Thanks in advance for your help.

  • 1
    $\begingroup$ How about $f(x)=e^x$ for irrational $x$, $f(x)=e^{-x}$ for rational $x$? This is continuous only at $x=0$. $\endgroup$ Jan 17 '14 at 16:46
  • $\begingroup$ @David Mitra, your function never takes the value $0$. The question is to find a bijection between $\mathbf{R}$ and $[0,+\infty)$. $\endgroup$
    – user121926
    Jan 17 '14 at 16:52
  • $\begingroup$ Oh, darn... (I saw "$\Bbb R^+$".) $\endgroup$ Jan 17 '14 at 16:54
  • $\begingroup$ Yes, Bourbaki decided that $0$ should be considered positive, and there are a lot of people who follow that convention. $\endgroup$
    – user121926
    Jan 17 '14 at 16:56
  • $\begingroup$ The first part of Julien's question, which he said he'd solved, is answered here: math.stackexchange.com/questions/8149/… $\endgroup$
    – user121926
    Jan 17 '14 at 16:57

As written, the first part isn't true. Consider $f(x)=e^x,$ for example.

Now, there are examples in which we do have infinitely-many discontinuities. One common approach is to "shift the problem to infinity," so to speak. Let's adapt the above function $f$ to do the trick. Define $g:\Bbb R\to\Bbb R^+$ by $$g(x)=\begin{cases}f(x-1) & \text{if }x\in\Bbb Z\\f(x) & \text{otherwise.}\end{cases}$$ Basically, we've started with the graph of a continuous bijection, then taken infinitely-many points on that graph and shifted them to the right by $1,$ thereby creating infinitely-many discontinuities, while making sure that we still have a bijective function.

Added: The above assumes that $\Bbb R^+$ refers to the positive real numbers. If it refers to the non-negative real numbers, then the first part is true, with the reasoning you mention being enough to do the trick. We can still adapt the function $f(x)$ to get a function of the desired sort, but we need to approach it a bit differently. The kicker is that $f$ is a function from $\Bbb R$ into the positive reals. We'll need to make one of our function values $0,$ but that leaves a hole in the range that must be filled if we're to have a bijection. So, we can fill that hole by making another adjustment, leaving another hole to be filled, and so on. One idea is to shift the function-value $1$ down to $0,$ shift the function-value $2$ down to $1,$ and in general, for positive integers $n,$ shift the function-value $n$ down to $n-1.$ Given any such $n,$ we have $f(x)=n$ precisely when $x=\ln n.$ Thus, we want our adapted function to be $$h(x)=\begin{cases}f(x)-1 & \text{if }x=\ln n\text{ for some positive }n\in\Bbb Z\\f(x) & \text{otherwise.}\end{cases}$$ Since we continue the shift/replace process indefinitely, all the holes are filled in, and it can be shown that $h$ is a bijection $\Bbb R\to\Bbb R^+,$ as desired.

  • $\begingroup$ It's clear to me that Julien means $[0,+\infty)$ when he writes $\mathbb{R}^{+}$. $\endgroup$
    – user121926
    Jan 17 '14 at 16:47
  • $\begingroup$ Oops! I'll bet you're right. $\endgroup$ Jan 17 '14 at 16:48

For the second part, it is easy to find a bijection between $\mathbf{R}$ and $(0,+\infty)$. (The exponential map is an example.) So you just need to find a bijection between $[0,+\infty)$ and $(0,+\infty)$. Define a map as the identity except on $\mathbf{N}$, and on $\mathbf{N}$ use $n \mapsto n + 1$.


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