Reflexive relations and the diagonal

In a first-year FPM workshop last year, I mentioned the diagonal in the Cartesian square of a set, and the connection with reflexive relations on the set. Relations R on a set S correspond to subsets M of S\times S in a standard way. And then the reflexive relations on S correspond to those subsets M of S\times S such that the diagonal is a subset of M.

I had a question on Piazza asking me to explain a bit more about what the diagonal is. Here is my reply.

Hi,
For a set S you can form the Cartesian square S \times S=\{(x,y):x,y \in S\,\}\,. For example \mathbb{R}\times \mathbb{R}= \{(x,y):x,y \in \mathbb{R}\}=\mathbb{R}^2\,. The diagonal (let's call it D) is a special subset of the Cartesian square S \times S defined by D=\{(x,y)\in S \times S:y=x\,\} = \{(x,x): x\in S\}\,. You can think of this as the graph of the identity map i_S:S\to S (where, for all x \in S, i_S(x)=x).
If S=\mathbb{R}, the diagonal is just the straight line through the origin in \mathbb{R}^2 given by the equation y=x.
If S is a finite set (for example, consider the special case S=\{1,5,9\}), then S \times S will also be a finite set (in our special case the Cartesian square has 3^2=9 elements) and the diagonal will be a finite set with the same number of elements as S (in our special case, the diagonal is just \{(1,1),(5,5),(9,9)\}).
Best wishes,
Dr Feinstein

Comments

Popular posts from this blog

A rule of thumb for when to use the Ratio Test

An introduction to the Hahn-Banach extension theorem: Part VI

Discussion of the proof that the uniform norm really is a norm