Open sets, closed sets, and continuity

Recall definition of open set: for every x ∈ U, there exists r > 0 such that B_{r}(x) is contained in U. Recall triangle inequality: |x + y| ≤ |x| + |y|. Proposition. The open ball B = {x ∈ ℝ^{n} | |x| < 1} is open.

Proof.Let x ∈ B. Then |x| < 1, and so ε = 1 − |x| > 0. We want to show that B_{ε}(x) ⊂ B. Suppose y ∈ B_{ε}(x). Then

|y| = |y − x + x| ≤ |y − x| + |x| < ε + |x| = 1,

and so y ∈ B. This shows that B_{ε}(x) ⊂ B, and since x was arbitrary, B is open. □

Recall equivalent definitions of closed set:

(1) For every convergent sequence {x_{n}} contained in X, the limit x_{0} is contained in X.

(2) X is closed if X^{c} is open.

Proposition. The closed ball \overline{B} = \{x ∈ ℝ^{n} | |x| ≤ 1\} is closed.

Proof.We will show that \overline{B}^{c} is open. Let x ∈ \overline{B}^{c}. Then |x| > 1, and so ε = |x| − 1 > 0. We will show that B_{ε}(x) is contained in \overline{B}^{c}.

To do this, we need a variant of the triangle inequality. Because we have both

|x| = |x − y + y| ≤ |x − y| + |y| which implies |x| − |y| ≤ |x − y|,  and
|y| = |y − x + x| ≤ |y − x| + |x| which implies |y| − |x| ≤ |y − x| = |x − y|,

we conclude

||x| − |y|| ≤ |x − y|.

This simply means that, in a triangle, the length of any one side is greater than the difference in lengths of the other two sides.

Now, let y ∈ B_{ε}(x). Then we have

|y| = |x − x + y| ≥ ||x| − |x − y|| > ||x| − ε| = 1 (because |x − y| < ε)

and therefore y ∈ \overline{B}^{c}. We conclude B_{ε}(x) ⊂ \overline{B}^{c}, and hence \overline{B}^{c} is open. By definition (2) of a closed set, \overline{B}^{c} is closed. □

Recall definition of continuity: f is continuous at x if either of the following equivalent conditions are satisfied:

(1) ∀ε > 0 ∃δ > 0 s.t. |x − y| < δ (i.e., y ∈ B_{δ}(x)) ⟹ |f(x) − f(y)| < ε.

(2) for all sequences {x_{n}} such that x_{n} → x, and for all ε > 0, there exists M ≥ 0 such that n ≥ M implies |f(x) − f(x_{n})| < ε.

Proposition. The function | · | : ℝ^{n} → ℝ (defined by x ↦ |x|) is continuous.

Proof.Choose x ∈ ℝ^{n}. We want to show, using definition (1) of continuity, that | · | is continuous at x. Let ε > 0, and set δ = ε. Suppose y ∈ ℝ^{n} satisfies |x − y| < δ. Then the triangle inequality implies that

||x| − |y|| ≤ |x − y| < δ = ε.

This shows that | · | is continuous at x, and since x was arbitrary, | · | is continuous on all of ℝ^{n}. □