A set can be described in many different ways.
Here are two different descriptions of a single set:
For instance, this set contains
(take and ) but does not contain
(the first component gives but that clashes with the third component,
similarly the first component gives 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,
and , are equal is to show mutual inclusion:
any member of is also in , and
any member of is also in .
To show that
we show first that and then that .
For the first half we must check that any vector
from is also in .
We first consider two examples to use them as models for the general argument.
If we make up a member of by trying and ,
then to show that it is in we need and such that
that is, this relation holds between and .
if we try and , then to show that the resulting
member of is in we need and such that
that is, this holds.
In the general case,
to show that any vector from is a member of we must show
that for any and there are appropriate and .
We follow the pattern of the examples; fix
and look for and such that
that is, this is true.
Applying Gauss' method
gives and .
This shows that for any choice of and there are appropriate
We conclude any member of is a member of 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 and there are appropriate
So fix and and solve for and :
shows that and .
Thus any vector from
is also of the right form for
Of course, sometimes sets are not equal.
The method of the prior example will help us see the relationship
between the two sets.
are not equal sets.
While is a subset of , it is a proper subset of because
is not a subset of .
To see that, observe first that given a vector from
we can express it in the form for — if
we fix and , we can solve for appropriate , , and :
shows that that any
can be expressed as a member of with
, , and :
But, for the other direction, the reduction
resulting from fixing , , and and looking for and