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I have set $A =[-3,0]$ and set $B =(1,4)$. I have to prove the equivalence of these sets by constructing a bijective function from A to B and from B to A. I don't understand how to construct the function from A to B, since i have to get rid of the -3 and 0, using the Hilbert's hotel paradox.

spaceisdarkgreen
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user1263607
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  • To be clear, the fact that it is $-3,0,1,4$ are irrelevant. You can chain together bijections to form a bijection, and going from $[-3,0]$ to $[0,1]$ is trivial, so this might as well have been asking to go from $[0,1]$ to $(0,1)$. The typical thing to do here is, as you say, something similar to hilbert's hotel and have the "extra" people displace some others who then displace the next people in line and so on. So, take the identity map for the majority of this, but let's have $0$ displace something and $1$ displace something else. – JMoravitz Dec 06 '23 at 18:24
  • Then, have what they displace continue to displace the next thing and so on... so for instance $0\mapsto \frac{1}{2}$, and $1\mapsto \frac{1}{3}$, then have $\frac{1}{2}\mapsto \frac{1}{4}$ and $\frac{1}{3}\mapsto \frac{1}{5}$ and in general $\frac{1}{n}\mapsto \frac{1}{n+2}$ and so on... – JMoravitz Dec 06 '23 at 18:25

1 Answers1

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Taking an inspiration from Hilbert's Hotel, you can do the following:

Start with a function $f_0: (-3,0) \to (1, 4)$ in the ordinary way; $f_0(x) = x+4$. You still need somewhere to send $-3$ and $0$ to define a function from $[-3, 0]$ to $(1,4)$, so you pick two infinite sequences in $(1,4)$ and "push them along," so to speak. We can use $1 \frac 12, 1 \frac 34, 1 \frac 78, 1 \frac {15}{16}, \dots$ as an example.

Define $f: [-3,0] \to (1, 4)$ to agree with $f_0$, except for the following:

We define $f(-3) = 1 \frac 12$. If this is all we do, then $f(-3) = f(-2 \frac 12)$, so we also define $f(-2\frac 12) = 1 \frac 34$. But then if we stopped there, we'd have $f(-2\frac 12) = f(-2\frac 14)$, so we define $f(-2\frac 14) = 1 \frac 78$, and so on. As in Hilbert's Hotel, we're moving guests to occupied rooms in sequence such that everyone only ever needs to move once and everyone finds a new room.

In general, we have the sequence $-3, -2 \frac 12, -2\frac 14, -2\frac 18, \dots$ in $[-3, 0]$. To find $f(x)$, we simply check if $x$ is in this sequence. If it's not, then we have $f(x) = x+4$ like we started with. If it is, we add 4 to the next element of the sequence; $f(-3) = -2\frac 12 + 4, \, f(-2\frac 12) = -2 \frac 14 + 4$, etc.

We can then do a similar trick for the point $0$, and we're done. Our function isn't continuous, but it turns out no bijection between an open interval and a closed interval can be continuous, so this is the best we can do.

DanishChef
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