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I want to define a piecewise-defined bijection $f: \Bbb R \to (\Bbb R$ \ $ \{1\})$ but I'm stuck.

This means that I must define $f(x)$ by cases: $f(x) = g_1(x)$ if $x \in J_1$, $f(x) = g_2(x)$ if $x \in J_2$,... where $J_1,J_2,...$ are intervals.

I don't know if this one works:

$f(x) = \frac{1}{x}$ if $x \in (-\infty, 0) \cup (0,1) \cup (1, 2) \cup (2, \infty)$, $f(x)=0$ if $x=0$, $f(x)= 2$ if $x=1$.

Edit

$f(x) = \frac{1}{1-x}$ if $x \in (-\infty, 1) \cup (1,\infty), \ f(x)=0$ if $x=1$.

Amanjo
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  • Is there a particular reason why you would need a piece-wise function? Then you could use $ \ f:x\to \frac 1{1-x}$. –  May 01 '14 at 23:39
  • There's no particular reason; it's just a problem I want to solve: find a bijective piecewise function $f:\Bbb R \to (\Bbb R$ \ $ {1})$. – Amanjo May 02 '14 at 00:01
  • @user129120 What's $f(1)$? – Dan Z May 02 '14 at 00:54
  • Are there any more restriction on intervals? Should they be half open intervals? How many intervals are allowed? – Sungjin Kim May 02 '14 at 00:55
  • @BiditAcharya The function is indeed a bijection between $\mathbb{R}-{1}$ and $\mathbb{R}-{0}$. – Sungjin Kim May 02 '14 at 00:56
  • There are no other restrictions. @Dan Zollers I edited my function. – Amanjo May 02 '14 at 01:48
  • @user129120: Great, now what's $f(2)$? [ Also, I think you wanted $f(1)=\frac12$; as it stands it's not injective because $f(1)=f(\frac12)$. ] – Eric Stucky May 02 '14 at 01:51
  • You are right. What about my new function? – Amanjo May 02 '14 at 02:02
  • @BiditAcharya But 1 is in the image of $f$... – Amanjo May 02 '14 at 03:59
  • @user129120 Wow, My bad! For some reason, I thought you were asking for a function $f:\mathbb R \setminus {1}\to \mathbb R$. However, Since the function I proposed is a bijection, its inverse is a function from $\mathbb R \to \mathbb R \setminus {1}$. Kind of a cheat but still! –  May 02 '14 at 04:08
  • The inverse function is $\frac {3}{2} ± \frac {\sqrt{x+4}}{2 \sqrt{x}}$. This function doesn't map $\Bbb R$: it has a real and an imaginary part. – Amanjo May 02 '14 at 04:16

1 Answers1

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Hint: I'm assuming that most of the functions you were planning on applying to your intervals were going to be continuous — or at worst have a couple of discontinuous points. In this case, you will need to use infinitely (thus countably) many intervals. This can be proven using the following argument:

Claim: Every bijection $f:\mathbb R\to\mathbb R\smallsetminus\{1\}$ has infinitely many points of discontinuity.

We argue by contrapositive: Assume that there are finitely many points of discontinuity $x_1<x_2<x_3<\cdots<x_n$. Define $x_0=-\infty$ and $x_n=\infty$. Then by definition, the restriction of $f$ to $$f^*: \left(\mathbb R\smallsetminus \{x_i\}_{i=1}^n\right) \to \left(\mathbb R\smallsetminus\left[\{1\}\cup\{x_i\}_{i=1}^n\right]\right)$$ is a continuous. By continuity, we know that the intervals $(x_k,x_{k+1})$ (with $0\leq k\leq n$) are mapped into intervals $I_k$. However, the domain is the union of $n+1$ intervals, but the codomain is a union of $n+2$ intervals, and so there is some interval in the codomain which is not in Im$(f^*)$ and hence not in Im$(f)$. Therefore, $f$ cannot be a bijection, which proves (by contrapositive) the claim.

Eric Stucky
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  • The function must be "picewise continuous". Does my last function work? – Amanjo May 02 '14 at 02:07
  • @user129120: Do you see that the claim is saying is that any piecewise continuous function needs infinitely many pieces? [This means that no, your last function does not work. Looking at the details, we see this is because $f(0)=1$.] – Eric Stucky May 02 '14 at 02:28
  • Ah, I see the wording was a bit confusing: the things which I was assuming were continuous in my hint were the functions you applied to the intervals, that is, the pieces. So really I was describing a piecewise continuous function, but I was trying to distinguish between the pieces and the regions of continuity, which in hindsight was a pointless nitpick. My apologies. – Eric Stucky May 02 '14 at 02:36