1.Prove that the intersection of two intervals is again an interval. Is the same true for unions?
Proof: Firstly, we need to prove that . It’s vacuously true. Assume, on the contrary, .
In other words, there must exist two elements x,y such that there exist at least one element z(where x<z<y) that is not in the set.
Here rasies a contradction, since there are no elements in the . Hence, our assumption is invalid, so .

Secondly, we come to prove that the intersection of two intervals is again an interval. Let A=(a,b)={x|a<x<b}, Let B=(c,d)={x|c<x<d}.
Case 1: . We’ve already proven that it’s a vacuous truth.
Case 2: = (min(a,c), min(b,d)). Define min(a,c) as the smallest number between a and c.
Hence, = (min(a,c), min(b,d)) is an interval.

Last but not least, we need to show that the unions of two intervals aren’t intervals. Since it’s easy to find two elements x,y any unions of two intervals such that there exist at least one element z(where x<z<y) that is not in the set.

Relavant: {x| x image} {x| x is the answer in the question}
Spurious: {x| x is the answer in the answer column} {x| x image}
Y: {x| x is the answer in the answer column} {x| x image}
Consider a scenario where a question specifies a property, such as stating ‘it is a painting.’ If an image, denoted as {x | x ∈ image}, lacks this property and is considered spurious by the definition, should it be labeled as ‘S’?