## 1.3 Exercise 10

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.)

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SOLUTION

(a)

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.

(b)

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:

Now,

$g(f(x)) = x$

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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