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I am supposed to construct a bijective function for the interval: \begin{align} I_2=\left(-\frac{\pi}{2} ,\frac{\pi}{2} \right] \longrightarrow \mathbb{R} \tag{Problem} \end{align} I first tried the easier case, i.e. \begin{align}f_1:I_1=\left(-\frac{\pi}{2} ,\frac{\pi}{2} \right) &\longrightarrow \mathbb{R} \\ x& \longmapsto \tan(x) \end{align} which is a bijection. Now I know that the composition of bijective functions is still a bijection. Which means that it should be possible to 'make room' for the missing point $\pi/2$. The following function: \begin{align}\phi : \mathbb{R} &\longrightarrow \mathbb{R \setminus}\lbrace 0 \rbrace \\ x & \longmapsto \begin{cases}x+1 \ \text{if} \ x \in \mathbb{N}_0 \\x \ \text{otherwise} \end{cases}\end{align} would be bijective, such that the composition $\phi \circ f_1: I_1 \rightarrow \mathbb{R}\setminus \lbrace 0 \rbrace $ is bijective, and as desired it now has room for a point that I can map to.

At this point I am not sure if my approach is correct because I can't find a function that would do the trick. Would I need to come up with another composition or is it enough to define a function that maps to the functions introduced above?

Update (in consideration of the answers given)

If I understand things correctly I can define: \begin{align}f_2: I_2 &\longrightarrow \mathbb{R} \\ x& \longmapsto \begin{cases}\phi(x) \ \text{for} \ x \in I_2 \\0 \ \text{for} \ x=\frac{\pi}{2} \end{cases} \end{align} Update 2 (Clarification required).

Define a new function to be equal to $\phi f_1$ over $I_1$ and have it map $π/2$ to $0$. As suggested (and upvoted) by @TBrendle.

If I do understand this correctly, then I need to map $x=\frac{\pi}{2}$ to $0$. However in this case it would make no sense to me to include $I_1$ in the domain, because $\pi/2$ is not in the domain, hence I don't see why I should include it in the codomain, however if I define: \begin{align}w: I_2 &\longrightarrow \mathbb{R} \\ (\phi\circ f_1)(x) & \longmapsto \begin{cases} (\phi \circ f_1) \ \text{if} \ x \in I_2 \\ 0 \ \text{if} \ x= \frac{\pi}{2} \end{cases} \end{align} This doesn't even look like a legitimate function to me anymore, since at $x=0$ the function evaluates to both $0$ and $1$.

Spaced
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    possible duplicate of http://math.stackexchange.com/questions/300815/show-that-open-segment-a-b-close-segment-a-b-have-the-same-cardinality and http://math.stackexchange.com/questions/28568/bijection-between-an-open-and-a-closed-interval – TBrendle Oct 21 '13 at 18:26
  • I also asked a related question a while back... http://math.stackexchange.com/questions/362933/creating-a-bijection-from-a-b-to-mathbb-r-that-is-visually-compelling – The Chaz 2.0 Oct 21 '13 at 19:58
  • You are deceiving yourself in an unhelpful way about what is required. You have described a function, it's just not presented in a single "formula" of the simplest type. – paul garrett Oct 21 '13 at 20:04
  • Maybe the update I have provided in my question shows a sufficient way how to handle a definition of such a function. – Spaced Oct 21 '13 at 20:59

3 Answers3

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Your approach is flawless. What were you worried about? Your function will be defined piecewise.

Added details:

Your new function is $$\psi : \left(-\frac{\pi}{2} \frac{\pi}{2}\right] \to \mathbb{R}$$

given by

$$ \psi(x) = \begin{cases} \phi(\tan x), & \text{for }-\frac{\pi}{2} < x< \frac{\pi}{2} \\ 0, & \text{for } x=\frac{\pi}{2}\\ \end{cases} $$ where $\phi$ is as given in the question. You can verify that the function $\psi$ has the specified domain and range and is a bijection.

TBrendle
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  • Thanks, it took me quite a while to get this far, however I need to find a bijection from $I_2 \rightarrow \mathbb{R}$ as suggested in the problem. So far I have introduced a bijection $ \phi \circ f_1$, it seems to like I am missing one final step that introduces another function which has all the characteristics as described above but maps from $I_2 \to \mathbb{R}$ – Spaced Oct 21 '13 at 18:34
  • Define a new function to be equal to $\phi f_1$ over $I_1$ and have it map $\frac{\pi}{2}$ to 0. Then you've done all the hard work already, you can see this is a bijection from your interval to the real line. – TBrendle Oct 21 '13 at 18:35
  • would that be $ \phi \circ f_1 \rightarrow (\phi \circ f_1) \in I_1$ otherwise $(\phi \circ f_1) (\frac{\pi}{2})=0$ ? – Spaced Oct 21 '13 at 19:00
  • If I could request for some additional information/clarification I would be very grateful, I summarized my attempts and misunderstandings in my post above @TBrendle. – Spaced Oct 22 '13 at 16:55
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Why not find a bijection from (-pi/2, pi/2] to the open interval and then compose with tangent?

Jonathan Aronson
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  • Any suggestions for this bijection? I'm not saying there is no such bijection (in fact, there has to be) $-$ but it's not at all easy to find. – TonyK Oct 21 '13 at 19:00
  • pi/2 goes to 0, 0 goes to 1/2, 1/2 goes to 2/3, 2/3 goes to 3/4 etc. seems easy to me. or am I missing something? – Jonathan Aronson Oct 21 '13 at 19:52
  • if you agree with me please get rid of your downvote :( – Jonathan Aronson Oct 21 '13 at 19:54
  • No. Your answer is incomplete without your comment. (And anyway a downvote is just a downvote.) – TonyK Oct 22 '13 at 16:07
  • It is a homework problem, so I thought I would let the student figure it out. After all the bijection is almost identical to the bijection $\phi$ defined in the post. I would be sour in your position too, so I understand your decision. – Jonathan Aronson Oct 22 '13 at 20:15
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This is a little kludgy, but I think it works. Let

$$\begin{align} f(x) &= {\pi/2-x\over x}\text{ for } 0\lt x\le\pi/2\cr &={x-\pi/4\over x+\pi/4}\text{ for } -\pi/4\lt x\le 0\cr &={4\over\pi}x+{1\over2}\text{ for } -3\pi/8\lt x\le-\pi/4\cr &={4\over\pi}x+{5\over4}\text{ for } -7\pi/16\lt x\le-3\pi/8\cr &=\text{etc.}\cr \end{align}$$

where the "etc." means you've chopped the interval $(-\pi/2,-\pi/4]$ into finer and finer halves converging on $-\pi/2$ and assigned each interval a linear function with image $(-1/2^n,-1/2^{n+1}]$ The idea is that $(0,\pi/2]$ maps to $[0,\infty)$, $(-\pi/4,0]$ maps to $(-\infty,-1]$, and the rest of the half-open intervals map to $(-1,-1/2]\cup(-1/2,-1/4]\cup(-1/4,-1/8]\cup\cdots=(-1,0)$.

Barry Cipra
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