Archive for February 13th, 2009

1.3 Exercise 10

February 13th, 2009 No comments

This problem acquaints us with the notion that two sets can be the same order type.  The requirement is that a function between the sets preserve order and that it be bijective.

"(a) Show that the map  f : (-1,1) \rightarrow \mathbb R of Example 9 is order preserving.

 f(x) = \frac{x}{1-x^2}

(b) Show that the equation  g(y) = \frac{2y}{[1+(1+4y^2)^{\frac{1}{2}}]}  defines a function  g : \mathbb R \rightarrow (-1,1) that is both a left and right inverse for  f ."

(Taken from Topology by James R. Munkres, Second Edition, Prentice Hall, NJ, 2000. Page 29.)




If the map is order preserving, then  x_1 < x_2 \Rightarrow f(x_1) < f (x_2) .  Now, we've got to be careful or we will not be thorough.  We have to assume that the set  (-1, 1) inherits the standard order of the reals.

Pick  x_1 = 0, x_2 but in  (0, 1) , and thus  0 < x_2 by the standard order of the reals.  We've got to proceed by contradiction.  Suppose  f(0) \geq f(x_2) .  This means  0 \geq \frac{x_2}{1-x_2^2} .  Since the denominator is positive, we can multiply both sides of the inequality without affecting the ordering (we take this as an axiomatic property of inequality), to obtain  0 \geq x_2 , a clear contradiction.  

Next pick  x_1 but in  (-1, 0) and  x_2 = 0 , so that  x_1 < 0 by the standard order of the reals.  Again proceeding by indirect proof, assume  f(x_1) \geq f(0) , which means  \frac{x_1}{1-x_1^2} \geq 0 .  Again the denominator is positive for any choice of  x_1 in the domain, so we can multiply without affecting the ordering.  In this case,  x_1 \geq 0 , contrary to our assumption.  

Now, by transitivity on zero, any  x_1 < x_2 with  x_1 < 0 and  0 < x_2 (but within the domain  (-1, 1) ) implies  f(x_1) < f(0), f(0) < f(x_2) \Rightarrow f(x_1) < f(x_2) . We have yet to check that any two positive choices  x_1, x_2 and two negative choices  x_1, x_2 (within the domain) are ordered under the image.

Pick two  x_1, x_2 > 0, x_1 < x_2 within  (0, 1) .  By indirect proof,


We know from hypothesis that  x_1 < x_2 \Rightarrow x_1 - x_2 < 0 .  Thus, by transitivity of the reals:


This last step is justified because, since both  x_1, x_2 are positive quantities, we can divide both sides by either one without affecting the ordering.  Now, the conclusion is contrary to our assumption that   x_1 < x_2 .  Thus  x_1 < x_2 \Rightarrow f(x_1) < f (x_2) must be true for two positive choices of  x_1, x_2 in the domain. 

Lastly, pick two   x_1, x_2 < 0, x_1 < x_2 within  (-1, 0) .  By indirect proof,


Let's make the negative sign explicit so that the elements  x_1, x_2 are positive and within  (0, 1) . But this in particular now means that  x_2 < x_1 \Rightarrow x_2 - x_1 < 0 .

 \frac{x_1}{x_1^2 - 1} \geq \frac{x_2}{x_2^2 - 1}

The denominators are both negative quantities.  In particular, when we multiply by one there will be an inequality reversal, but multiplying then by the other preserves the inequality.  Thus:


By transitivity on zero, 


This last bit is contrary to our assumption, and we are done.

Therefore  f is order preserving. 


First we observe in particular that any  y works (negative or otherwise) because the squaring of  y makes  g(y) 's denominator positive  \forall y .  So there are no discontinuities and the domain of  g is all of  \mathbb R .  Secondly, we observe that if we pick  y a very large number,  1+4y^2 \approx 4y^2 and the denominator becomes approximately  1+2y \approx 2y .  Thus  g(y) < 1 for  y a very large number.  Arguing similarly,  g(y) > -1 for  y a very large negative number.  The image of  g(y) is therefore  (-1,1) .

We now show that  g(y) is a right inverse by calculating  f(g(y)) = y .  Then we show  g(y) is a left inverse by calculating   g(f(x)) = x .

 f(g(y)) = y

Just for fun, rearrange  g(y) as:





 g(f(x)) = x


The fact that  f is both order preserving and bijective means that  (-1, 1) and  \mathbb{R} are the same order type.