Stepan Malkov

Week 2

Open and Closed Sets

We review the definitions of open and closed sets.
Definition:
A subset A \subset X of a metric space is said to be open if it contains none of its boundary points, i.e \partial A \cap A = \varnothing. A is said to be closed if it contains all of its boundary points, i.e. \partial A \subseteq A.
Proposition:
The following are equivalent definitions of an open set:
  1. \partial A \cap A = \varnothing.
  2. For any a \in A, there exists an r>0 such that the open ball B(a,r) \subset A.
  3. X \setminus A is closed.
Proof
(1) \implies (2): Let a \in A. Then, a \not \in \partial A, i.e. a \in \text{int}(A). But this means precisely that there exists an r > 0 such that B(a,r) \subset A.
(2) \implies (3): Let x \in \partial (X \setminus A). Then, for any r>0, B(x,r) \cap (X \setminus A) \not = \varnothing, so x \not \in A. But then, x \in X \setminus A, i.e. X \setminus A is closed.
(3) \implies (1): let a \in \partial A. Then, for any r>0, B(a,r) \cap (X \setminus A) \not = \varnothing, i.e. a \in \partial (X \setminus A). But since X \setminus A is closed, this implies that a \in X \setminus A, so \partial A \cap A = \varnothing.
Proposition:
A is closed if and only if for any sequence (a_n)_{n=1}^\infty \in A such that \lim_{n \to \infty} a_n = x \in X, it follows that x \in A.
Proof:
Suppose A is closed, and let (a_n)_{n=1}^\infty \in A be a sequence such that a_n \to x \in X. Then, either x \in A,x \in \partial A, or x \in \text{ext}(A). If x \in A, we are done. If x \in \partial A, since \partial A \subseteq A, we are also done. If x \in \text{ext}(A), there exists an r>0 such that B(x,r) \cap A = \varnothing. But this contradicts the fact that a_n \to x, since for any r>0 one can find an N such that d(a_n,x) < r, i.e. a_n \in B(x,r) for n \geq N. Conversely, suppose that a_n \to x \in X implies x \in A. Let y \in \partial A. Want to show y \in A. But for any r>0, B(y,r) \cap A \not = \varnothing, i.e. can find an element a_n \in B(y,\frac{1}{n}). Then, d(a_n,y) < \frac{1}{n} \to 0, so a_n \to y. But this implies y \in A. Thus, A is closed.
Proposition:
An arbitrary union (finite intersection) of open sets is open. An arbitrary intersection (finite union) of closed sets is closed.
Examples:
  1. Any closed interval is a closed set in \mathbb{R} and any open interval is an open set in \mathbb{R}.
  2. D = \{(x,y) \in \mathbb{R}^2: x^2+y^2 \leq 1\} is closed in \mathbb{R}^2.
  3. Let (X,d) be any set with the discrete metric. Then, any subset A \in \mathbb{X} is closed and open.
  4. Let C([a,b]) be the space of continuous functions on the interval [a,b] with the metric d(f,g) = \sup_{x \in [a,b]} |f(x)-g(x)| (exercise to verify this is inded a metric). Then, the set

        \[X = \{f \in C([a,b]): f(a) = 0\}\]

    is closed but not open, for g(x) = x^2,

        \[Y = \{f \in C([a,b]): \sup_{x \in [a,b]}|f(x)-x^2|<1\} = B(g,1)\]

    is open but not closed,

        \[Z = \{f \in C([a,b]): f(a)<f(b)\}\]

    is neither closed nor open.
  5. In l^\infty = \{(a_n)_{n=1}^\infty: \sup_{n} |a_n| <\infty\}, the space of bounded sequences, the subset c_0 =\{(a_n)_{n=1}^\infty: \lim_{n \to \infty} a_n = 0\}\subset l^\infty is closed but not open.
As seen by the previous example, note that a set being not closed does not imply that it is open. Thus, sets can be closed, open, both, or neither. An example of a set in \mathbb{R} that is neither closed nor open in [-2,1). However, the only sets in a subset X \subset \mathbb{R}^n with the induced metric that are both closed and open are the empty set \varnothing (trivially), and the entire subspace X.
Definition:
Let (X',d) \subset (X,d) be a subspace with the induced metric. Then, A' \subset X' is open (closed) in X' if there exists an open (closed) set A \subset X such that A \cap X' = A'.
Example:
Take [0,1) \subset \mathbb{R} with the induced metric. Then, [0,\frac{1}{2}] is closed but not open, [0,\frac{1}{2}) is open but not closed, and [\frac{1}{4},\frac{3}{4}) is neither open nor closed. The only sets in [0,1) that are both open and closed are [0,1) and \varnothing.