Section 3: Problem 3 Solution
Working problems is a crucial part of learning mathematics. No one can learn topology merely by poring over the definitions, theorems, and examples that are worked out in the text. One must work part of it out for oneself. To provide that opportunity is the purpose of the exercises.
James R. Munkres
Here is a "proof that every relation
that is both symmetric and transitive is also reflexive: "Since
is symmetric,
implies
. Since
is transitive,
and
together imply
, as desired." Find the flaw in this argument.
To prove that a relation
is reflexive, we need to prove that for every
,
. What the proof above does, it shows that for some
such that there happens to be
such that
, we can show that
must hold. But what if there is
such that there is no
such that either
or
holds? Then the proof cannot be applied to such
to prove that
.
The easiest example is the empty relation on a non-empty set. It is (vacuously) symmetric and transitive. But, of course, it is not reflexive. Given any
, there is just no
such that either
or
holds, hence, the proof does not work.