This subsection is optional. Later material will not require the work here.
A set can be described in many different ways. Here are two different descriptions of a single set:
For instance, this set contains
(take z = 5 and w = 5 / 2) but does not contain
(the first component gives z = 4 but that clashes with the third component, similarly the first component gives w = 4 / 5 but the third component gives something different). Here is a third description of the same set:
We need to decide when two descriptions are describing the same set. More pragmatically stated, how can a person tell when an answer to a homework question describes the same set as the one described in the back of the book?
Sets are equal if and only if they have the same members. A common way to show that two sets, S_{1} and S_{2}, are equal is to show mutual inclusion: any member of S_{1} is also in S_{2}, and any member of S_{2} is also in S_{1}.^{[1]}
To show that
equals
we show first that and then that .
For the first half we must check that any vector from S_{1} is also in S_{2}. We first consider two examples to use them as models for the general argument. If we make up a member of S_{1} by trying c = 1 and d = 1, then to show that it is in S_{2} we need m and n such that
that is, this relation holds between m and n.
Similarly, if we try c = 2 and d = − 1, then to show that the resulting member of S_{1} is in S_{2} we need m and n such that such that
that is, this holds.
In the general case, to show that any vector from S_{1} is a member of S_{2} we must show that for any c and d there are appropriate m and n. We follow the pattern of the examples; fix
and look for m and n such that
that is, this is true.
Applying Gauss' method
gives n = (5 / 11)c − (3 / 11)d and m = (4 / 11)c + (2 / 11)d. This shows that for any choice of c and d there are appropriate m and n. We conclude any member of S_{1} is a member of S_{2} because it can be rewritten in this way:
For the other inclusion, , we want to do the opposite. We want to show that for any choice of m and n there are appropriate c and d. So fix m and n and solve for c and d:
shows that d = (5 / 2)m − 2n and c = (3 / 2)m + n. Thus any vector from S_{2}
is also of the right form for S_{1}
Of course, sometimes sets are not equal. The method of the prior example will help us see the relationship between the two sets. These
are not equal sets. While P is a subset of R, it is a proper subset of R because R is not a subset of P.
To see that, observe first that given a vector from P we can express it in the form for R— if we fix x and y, we can solve for appropriate m, n, and p:
shows that that any
can be expressed as a member of R with m = x, n = 2x, and p = y:
Thus .
But, for the other direction, the reduction resulting from fixing m, n, and p and looking for x and y
shows that the only vectors
representable in the form
are those where 0 = m + (1 / 2)n. For instance,
is in R but not in P.
Decide if the vector is a member of the set.
Produce two descriptions of this set that are different than this one.
One easy thing to do is to double and triple the vector:
Show that the three descriptions given at the start of this subsection all describe the same set.
Instead of showing all three equalities, we can show that the first equals the second, and that the second equals the third. Both equalities are easy, using the methods of this subsection.
Show that these sets are equal
and that both describe the solution set of this system.
That system reduces like this:
showing that w = 1, y = 4 and x = 2 − z.
Decide if the sets are equal.
For each item, we call the first set S_{1} and the other S_{2}.
