1.3 Exercise 10

February 13th, 2009 Leave a comment Go to 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.

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