Cartesian squares and ordered pairs

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Here is another question and answer from my FPM Piazza forum last autumn.

Question: Suppose that \(S\) is a set with two elements, say \(S=\{a,b\}\). When looking at elements of the Cartesian square \(S \times S\), are \((a,b)\) and \((b,a)\) the same element, or are they different elements? Does \(S \times S\) have four different elements, or only three?

My answer:

Hi,
The key term in the definition of Cartesian squares, and generally Cartesian products, is "ordered pair". When you use standard round brackets in this way, the order does matter. You have specified a first coordinate and a second coordinate.
For example, if you work in \(\mathbb{R} \times \mathbb{R} = \mathbb{R}^2\), the point \((1,0)\) (which lies on the x-axis) is different from the point \((0,1)\) (on the y-axis).

Many of the sets \(S\) we have looked at are subsets of \(\mathbb{R}\), and this results in \(S \times S\) being a subset of \(\mathbb{R}^2\). When this happens, you can often think of points in our Cartesian square as being points in 2-dimensional space.

For example, if \(S=\{1,3\}\), then \(S \times S\) has four different points, which you can think of as being points in \(\mathbb{R}^2\) that are at the corners of a square:
         \(S \times S = \{(1,1), (1,3), (3,1), (3,3)\}\,.\)
On the other hand, if \(S\) is the closed interval \([0,1]=\{x \in \mathbb{R}: 0 \leq x \leq 1\}\), then \(S \times S\) really is the "unit square" \(\{(x,y) \in \mathbb{R}^2: 0 \leq x \leq 1, 0 \leq y \leq 1\,\}\).

Best wishes,
Dr Feinstein

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