Question

Decide if the following statements are true or false. Explain. $$ \left\\{(x, y) \in \mathbb{R}^{2}: x^{2}-x=0\right\\} \subseteq\left\\{(x, y) \in \mathbb{R}^{2}: x-1=0\right\\} $$

Short answer

The statement is true, the set defined by \(\left\{(x, y) \in \mathbb{R}^{2}: x^{2}-x=0\right\}\) is indeed a superset of \(\left\{(x, y) \in \mathbb{R}^{2}: x-1=0\right\}\). This is because the first set includes coordinates for which \(x\) is either 0 or 1, while the second set includes only those with \(x = 1\). Hence, all elements of the second set also exist in the first set.

Step-by-step solution

Define the conditions

The first requirement is to express the equations given in the conditions as sets of coordinates on the plane. The given conditions are: \(x^2 - x = 0\) and \(x - 1 = 0\). The first one can be factorized to \(x (x - 1) = 0\), which yields two potential values for x, \(x = 0\) and \(x = 1\). The second condition implies \(x = 1\).

Build the two sets

Using the values derived from the two conditions, it's possible to define the two sets. The first set, defined by \(x^2 - x = 0\), will include all coordinates for which \(x\) is either 0 or 1, regardless of the value of y, i.e., \(\{(x, y) | x \in \{0, 1\}, y \in \mathbb{R}\}\). For the second set, only one value of \(x\) is valid, namely \(x = 1\), regardless of y. Therefore, it will consist of all coordinates for which \(x = 1\) and \(y\) can be any real number, i.e., \(\{(x, y) | x = 1, y \in \mathbb{R}\}\).

Compare the sets

Now, the subset relationship can be verified. For the first set, any coordinate with \(x = 0\) or \(x = 1\) is included, while the second set solely includes coordinates for which \(x = 1\). Hence, all elements of the second set exist in the first set: \(\{(x, y) | x = 1, y \in \mathbb{R}\} \subseteq \{(x, y) | x \in \{0, 1\}, y \in \mathbb{R}\}\), which corresponds to the original question, \(\left\{(x, y) \in \mathbb{R}^{2}: x^{2}-x=0\right\} \subseteq\left\{(x, y) \in \mathbb{R}^{2}: x-1=0\right\}\).