Question

Let A be a non-empty subset of R, not necessarily closed. define g:R→R by g(x)=inf{∣x−a∣:a∈A}.

(a) Prove g is continuous,
(b) Assuming now A closed, prove g(x)=0 if and only if x∈A. Is this result true if A is not closed?

Answer 1

The function g(x) = inf is continuous. Assuming A is closed, g(x) = 0 if and only if x ∈ A. This is not true if A is not closed.

Step-by-step explanation:

To prove that the function g(x) = infx-a is continuous, we can show that for any given ε > 0, there exists a δ > 0 such that |x - c| < δ implies |g(x) - g(c)| < ε for any x, c ∈ R.

To prove (a), let's choose δ = ε. Then, for |x - c| < δ, we have:

|g(x) - g(c)| = |infx - a - inf: a∈A| ≤ |x - c| < ε.

To prove (b), assuming A is closed, we can show that g(x) = 0 if and only if x ∈ A.

First, let's prove that if x ∈ A, then g(x) = 0. Since x ∈ A, we can choose a sequence (a_n) in A such that a_n → x as n → ∞. Therefore, for any ε > 0, there exists N such that |x - a_n| < ε for n ≥ N. This implies that g(x) = inf = 0.

Now, let's prove that if g(x) = 0, then x ∈ A. Since g(x) = 0, there exists a sequence (a_n) in A such that |x - a_n| → 0 as n → ∞. This implies that a_n → x as n → ∞. Therefore, x ∈ A.

This result is not true if A is not closed, since there can be points in A's closure that are not in A.