In set-builder notation, is the set of all x with 2 ≤ x ≤ 5, x = 10 the same as the interval from 2 to 5 together with the single point 10?

Step-by-step solution

1. 1

   State the convention for the comma. In set-builder notation everything after the bar is a list of conditions the element must satisfy simultaneously. The comma is read as "and":

   $\{x \mid P(x),\ Q(x)\} = \{x \mid P(x) \ \text{and} \ Q(x)\}$

   This is a fixed convention, not a stylistic choice — the same way commas work in $\{(x,y) \mid x > 0,\ y > 0\}$, the open first quadrant.

2. 2

   Apply the convention to the given set. Reading the comma as "and" turns the description into

   $2 \le x \le 5 \quad \text{and} \quad x = 10$

   No number satisfies both: $x = 10$ forces $x = 10$, but $10 > 5$, so the interval condition fails.

3. 3

   Conclude that the set is empty. The two conditions are contradictory, therefore

   $\{x \mid 2 \le x \le 5,\ x = 10\} = \varnothing$

   So it is not the intended set — it is the empty set, the most wrong answer possible rather than a near miss.

4. 4

   Write the intended set correctly. For a union you must use "or", which is a single condition built from a disjunction:

   $\{x \mid 2 \le x \le 5 \ \text{or} \ x = 10\}$

   Equivalently, take the union of two separately described sets:

   $\{x \mid 2 \le x \le 5\} \cup \{10\} = [2,5] \cup \{10\}$

5. 5

   Check membership on test values. In the corrected set: $x = 3$ satisfies the first clause, so $3$ belongs; $x = 10$ satisfies the second, so $10$ belongs; $x = 7$ satisfies neither, so $7$ does not. That is exactly the membership pattern of $[2,5] \cup \{10\}$, so the two descriptions agree.

6. 6

   Remember the pairing rule. Comma (and intersection $\cap$) go together and shrink a set; the word "or" (and union $\cup$) go together and grow it. Whenever a description is a list of separate pieces, you need "or" or a $\cup$, never a comma.