## 1.2 Exercise 2

"Let and let and for and . Show that preserves inclusions, unions, intersections, and differences of sets:

(a)

(b)

(c)

(d)

Show that preserves inclusions and unions only:

(e)

(f)

(g) ; show that equality holds if is injective.

(h) ; show that equality holds if is injective."

(Taken from *Topology* by James R. Munkres, Second Edition, Prentice Hall, NJ, 2000. Page 20.)

NB: A word about notation. in this context is the preimage of the subset of , and can therefore return a set (i.e., it is not a function), as will be seen presently.

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SOLUTION

(a)

Case

Pick such that the preimage in the domain is defined ( may not be surjective or even one-to-one). Apply on such, and then . Since , and it also belongs to by (proper) inclusion, . Now pick , with defined preimage in . Apply . Then . Thus .

Case

It is easy to see that, by arguing similarly, .

Thus .

(b)

Case

Pick a , for which the preimage is deifned in . . Pick . . We have covered every in , which preimage is .

Case

Pick a , well defined in . This element, under , belongs to and is therefore in . Since , it also belongs to , and has a preimage in .

Pick , well defined in . . Such also belongs to , and therefore .

We've now covered every in , and we are done.

Thus .

(c)

Case

Pick with a defined preimage in . Since it is in , the preimage lies in .

Such a belongs to , and so its preimage is in . It also lies in , and so its preimage lies also in . In other words, its preimage lies in .

Case

Pick , with defined preimage in . Pick with defined preimage in . take the intersection .

A mapping to the intersection has a preimage that maps both to and . In other words, such a and , or , so the preimage of lies also in .

Thus .

(d)

Case

Pick , with well defined preimage in . Since , its preimage lies in . But it cannot lie in , so therefore its preimage cannot lie in . In other words, the preimage of such a is in .

Case

Pick a with preimage in . Such a has a preimage in , but not in . This means that such a must belong to , but not : , or, in other words, . Such a has a preimage in .

Thus .

(e)

Case

Pick and apply . Such . By inclusion, , and therefore . Pick but . Apply . Then but . Thus .

Case

Pick (and ). Then and . Thus .

Therefore .

(f)

Same as (b) but arguing form the domain side.

(g)

Case

Pick (with image ). Now means that , and means , so such an 's image is in .

Case if is injective

is injective means that there is a unique that yields , in other words . So pick . The that was picked necessarily belongs to in order that be in (i.e., there isn't another outside or inside whose image is , due to injectivity), but it also must belong to so that be in (same argument here). Therefore , and the image of this belongs to .

If the argument above isn't crystal clear, the contrapositive is also insightful: if , is not injective. Pick that is not also in the image . In particular this means and such , say. But such a and , in other words, and this means that such an and , or . The only way to resolve this apparent contradiction is by letting , and is not injective.

(h)

Case , or

Pick . Since belongs to the image , there exist elements collectively , and focus on them (other elements are not of interest because they are not in ). Since does not belong to , these . Such exactly belong to , and therefore to .

Next we want to show that if is injective, . Since all s uniquely map into , pick any , and apply . Such will lie in . Since , and , this implies and .

The statement "since belongs to the image then belongs to " is false.

may not be an element of but may still be equal to

Sorry, I'm not sure what you mean. You wrote: " may not be an element of but may still be equal to " makes little sense to me... I may be missing something. Perhaps what you meant is that an element outside of can still map to . If this is the case, then you are right but we are opting for those elements which definitely belong to , because . I'm correcting the wording in the original post a bit to make it clearer.