Bijection between Sets(again)

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October 20th 2009, 08:48 AM
p00ndawg

Bijection between Sets(again)

Show that the following pairs of sets S and T are equinumerous by finding a specific bijection between the sets in each pair.

S = [0,1] and T = [0,1)
October 20th 2009, 09:05 AM
aman_cc

Quote:

Originally Posted by p00ndawg

Show that the following pairs of sets S and T are equinumerous by finding a specific bijection between the sets in each pair.

S = [0,1] and T = [0,1)

How about this f: S -> T
for x<1/2 f(x)=x
for x>=1/2 f(x) = 3/2-x

I guess it works

Sorry - This seems wrong !!
October 20th 2009, 09:27 AM
Plato

Quote:

Originally Posted by p00ndawg

Show that the following pairs of sets S and T are equinumerous by finding a specific bijection between the sets in each pair.

S = [0,1] and T = [0,1)

Try $f(x) = \left\{ {\begin{array}{rl} {x,} & {~x \ne \frac{1} {n},\,\left( {\forall n \in \mathbb{Z}^ + } \right)} \\ {\frac{1} {{n + 1}},} & {x = \frac{1} {n},\,\left( {\exists n \in \mathbb{Z}^ + } \right)} \\ \end{array} } \right.$
October 20th 2009, 09:40 AM
aman_cc

Quote:

Originally Posted by Plato

Try $f(x) = \left\{ {\begin{array}{rl} {x,} & {~x \ne \frac{1} {n},\,\left( {\forall n \in \mathbb{Z}^ + } \right)} \\ {\frac{1} {{n + 1}},} & {x = \frac{1} {n},\,\left( {\exists n \in \mathbb{Z}^ + } \right)} \\ \end{array} } \right.$

Plato - Thanks.

Please validate my understanding. You have used a technique similar to the argument used in Hotel Infinity Paradox - To make room available ask every occupant to shift to the next room.

Consider F: [0,1]->[0,1]

If I remove finite points from the range - I can use your argument to construct a bijection. Correct?
Say if range is (0,1), I will map 1/n -> 1/(n+2) and 0 -> 1/2

Correct plz?
October 20th 2009, 09:44 AM
p00ndawg

Quote:

Originally Posted by Plato

Try $f(x) = \left\{ {\begin{array}{rl} {x,} & {~x \ne \frac{1} {n},\,\left( {\forall n \in \mathbb{Z}^ + } \right)} \\ {\frac{1} {{n + 1}},} & {x = \frac{1} {n},\,\left( {\exists n \in \mathbb{Z}^ + } \right)} \\ \end{array} } \right.$

hmm, i dont understand this,im not seeing the bijection.

why the 1/(n+1)?
October 20th 2009, 09:51 AM
Plato

Quote:

Originally Posted by p00ndawg

hmm, i dont understand this,im not seeing the bijection.

You can prove that $f$ is an injection and a surjection for $[0,1]\mapsto [0,1)$ .

Have you tried?
October 20th 2009, 10:14 AM
p00ndawg

what do you mean by when x != (1/n)?

im working on the injection on surjection right now, but when x != (1/n) is that for everything in S = [0,1] ?
October 20th 2009, 10:49 AM
Plato

Quote:

Originally Posted by p00ndawg

what do you mean by when x != (1/n)?

im working on the injection on surjection right now, but when x != (1/n) is that for everything in S = [0,1] ?

Here are some examples,
$f(1)=\frac{1}{2},~f(\frac{1}{2})=\frac{1}{3},~f(\f rac{2}{3})=\frac{2}{3},~f(\frac{\pi}{4})=\frac{\pi }{4}$
October 20th 2009, 01:27 PM
p00ndawg

Quote:

Originally Posted by Plato

Here are some examples,
$f(1)=\frac{1}{2},~f(\frac{1}{2})=\frac{1}{3},~f(\f rac{2}{3})=\frac{2}{3},~f(\frac{\pi}{4})=\frac{\pi }{4}$

Okay! thank you! I think im finally getting it.

I just have one more question, when im looking at these two sets how would I know to
separate it into a function like that?

is it just practice? or is there anything from that set that is setting off some alarms in your head?

thanks a bunch!