**READING: I §3 6**
Back to reading**:**
I §3 5

*begin quote ita***"**

*CHAPTER I* [:]
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Notions from Set Theory
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These pages assume that the reader has the book and is reading along. topic index contents

Please read these notes from the beginning to here and then paragraph 6 of §3.

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ita-1-3-6-pg9
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If *f* : *X* → *Y* and *g* : *Y* → *Z* are functions, one can define the *composition of f and g*, or *composed function*, a function from *X* into *Z* by associating to each element of *X* and element of *Z* in the obvious way: given an element of *X*, one first uses *f* to get an element of *Y*, then one uses *g* to get from this last element an element of *Z*. The composed function is usually denoted *g* *f*, so that we have *g* *f* : *X* → *Z*, with ( *g* *f* )(*x*) = *g*( *f* (*x*)) for each .

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This composition of functions is a very simple idea, but I find that it is also confusing. We mention the function *f* before the function *g* and we call it the composition of *f* and *g*, but the symbol *g* *f* has *g* first. This last fact, however, is a memory device for the composition's value: ( *g* *f* )(*x*) = *g*( *f* (*x*)). The parentheses in this expression will also be a source of confusion for those who are not familiar with standard mathematical notation. The first set of parentheses indicate that *g* *f* is the symbol or name of the composed function, just as *f* and *g* are the names or symbols of functions. All three other pairs of parentheses enclose the arguments of the functions. Finaly, I would like to look at problem 6 on page 12:

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ita-1-Problem-6-pg12
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Prove that if *f* : *X* → *Y*, *g* : *Y* → *Z* and *h* : *Z* → *W* are functions, then *h* ( *g* *f* ) = ( *h**g* ) *f*.

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Let the proof be an exercise, and please see the example here.