## 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 of Example 9 is order preserving.

(b) Show that the equation defines a function that is both a left and right inverse for ."

(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 . Now, we've got to be careful or we will not be thorough. We have to assume that the set inherits the standard order of the reals.

Pick but in , and thus by the standard order of the reals. We've got to proceed by contradiction. Suppose . This means . 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 , a clear contradiction.

Next pick but in and , so that by the standard order of the reals. Again proceeding by indirect proof, assume , which means . Again the denominator is positive for any choice of in the domain, so we can multiply without affecting the ordering. In this case, , contrary to our assumption.

Now, by transitivity on zero, any with and (but within the domain ) implies . We have yet to check that any two positive choices and two negative choices (within the domain) are ordered under the image.

Pick two within . By indirect proof,

We know from hypothesis that . Thus, by transitivity of the reals:

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

Lastly, pick two within . By indirect proof,

Let's make the negative sign explicit so that the elements are positive and within . But this in particular now means that .

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 is order preserving.

(b)

First we observe in particular that any works (negative or otherwise) because the squaring of makes 's denominator positive . So there are no discontinuities and the domain of is all of . Secondly, we observe that if we pick a very large number, and the denominator becomes approximately . Thus for a very large number. Arguing similarly, for a very large negative number. The image of is therefore .

We now show that is a right inverse by calculating . Then we show is a left inverse by calculating .

Just for fun, rearrange as:

Now,

The fact that is both order preserving and bijective means that and are the same *order type*.