Is the Continuum Hypothesis Valid if the Interval is Wrapped into a Semi-Circle?

[TheAlkemist]

I was talking to a friend recently about the nature of infinity and Georg Cantor's Continuum Hypothesis and I recanted a statement from a Youtube video I watched:

"any interval of the continuum, now matter how small, has the same size as the contiuum itself"


from the 6:54 mark...

https://www.youtube.com/watch?v=<object width="480" height="385"><param name="movie" value="http://www.youtube.com/v/WihXin5Oxq8&hl=en_US&fs=1&"></param><param name="allowFullScreen" value="true"></param><param name="allowscriptaccess" value="always"></param><embed src="http://www.youtube.com/v/WihXin5Oxq8&hl=en_US&fs=1&" type="application/x-shockwave-flash" allowscriptaccess="always" allowfullscreen="true" width="480" height="385"></embed></object>

The mapping of the interval from A to B onto the continuum (through 1-to-1 correspondence pairing of elements) was done by first "wrapping up" the interval into a semi-circle and then pairing.

Now my friend thinks this is cheating. That once you bend the interval line (change the geometry) you have effectively changed the property of the line and therefor can't relate it to the continuum.

How do I explain to him that this isn't cheating?

Thanks in advance!

For the record, I'm not a mathematician so please forgive my ignorance :(

 

[Landau]

This probably shouldn't be in Topology & Geometry.

Now my friend thinks this is cheating. That once you bend the interval line (change the geometry) you have effectively changed the property of the line and therefor can't relate it to the continuum.

Well, it is of course an intuitive picture of what's going on. If you want to prove it, you should give the bijections explicitly. But ask him: if you have a piece of string of some length L, and you bend the string, will the length change? A bijection between the interval (0,pi) and a half circle can easily be given explicitly: send t in (0,pi) to e^{it}=(cos t, sin t).

If he just wants a solid proof of the whole thing:
There is a bijection
$$(0,1)\to (a,b)$$ given by $$x\mapsto a+(b-a)x$$
and a bijection
$$(0,1)\to \mathbb{R}$$ given by $$x\mapsto \tan\left(\pi(\x-\frac{1}{2})\right)$$.

Hence the composition
$$(a,b)\to\mathbb{R}$$ given by $$y\mapsto \tan\left(\pi\left(\frac{y-a}{b-a}-\frac{1}{2}\right)\right)$$
is a bijection, so $$|(a,b)|=|\mathbb{R}|$$.

 

[TheAlkemist]

This probably shouldn't be in Topology & Geometry.
Well, it is of course an intuitive picture of what's going on. If you want to prove it, you should give the bijections explicitly. But ask him: if you have a piece of string of some length L, and you bend the string, will the length change? A bijection between the interval (0,pi) and a half circle can easily be given explicitly: send t in (0,pi) to e^{it}=(cos t, sin t).

If he just wants a solid proof of the whole thing:
There is a bijection
$$(0,1)\to (a,b)$$ given by $$x\mapsto a+(b-a)x$$
and a bijection
$$(0,1)\to \mathbb{R}$$ given by $$x\mapsto \tan\left(\pi(\x-\frac{1}{2})\right)$$.

Hence the composition
$$(a,b)\to\mathbb{R}$$ given by $$y\mapsto \tan\left(\pi\left(\frac{y-a}{b-a}-\frac{1}{2}\right)\right)$$
is a bijection, so $$|(a,b)|=|\mathbb{R}|$$.

Thanks for this. :)