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Week 2 Discussion Notes

Table of Contents

Metric Space Topology

Open Balls

Definition

Let (X,d)\p{X, d} be a metric space. If x∈Xx \in X and r>0r > 0, then the open ball of radius rr centered at xx is

B(x,r)={y∈X∣d(x,y)<r}.B\p{x, r} = \set{y \in X \mid d\p{x, y} < r}.

When talking about topology (i.e., open sets, closed sets, etc.), open balls are like the "atoms." They're going to be the main object used in the definitions and concepts of this section.

Definition

Let (X,d)\p{X, d} be a metric space and E⊆XE \subseteq X be a subset.

  1. The interior of EE, denoted Int⁡E\Int{E}, is the set of points x∈Xx \in X where there exists r>0r > 0 such that B(x,r)⊆EB\p{x, r} \subseteq E.
  2. The exterior of EE, denoted Ext⁡E\Ext{E}, is the set of points x∈Xx \in X where there exists r>0r > 0 such that B(x,r)⊆X∖EB\p{x, r} \subseteq X \setminus E.
  3. The boundary of EE, denoted ∂E\bd{E}, is the set of points that are not in Int⁡E\Int{E} or Ext⁡E\Ext{E}.
Example 1.

(Anything dashed is not included in the set.)

Open and Closed Sets

Definition

Let (X,d)\p{X, d} be a metric space and E⊆XE \subseteq X be a subset.

  1. EE is open if for every x∈Ex \in E, there exists r>0r > 0 such that B(x,r)⊆EB\p{x, r} \subseteq E.
  2. EE is closed if its complement X∖EX \setminus E is open.

One thing to note that is open is not the opposite of closed.

Example 2.

Let (X,d)=(R,∣ ⋅ ∣)\p{X, d} = \p{\R, \abs{\:\cdot\:}}.

  1. E=(0,1)E = \p{0, 1} is an open set, but not closed.
    • E=(0,1)=B(12,12)E = \p{0, 1} = B\p{\frac{1}{2}, \frac{1}{2}}, so it's an open set. On the other hand, its complement is R∖E=(−∞,0]∪[1,∞)\R \setminus E = \poc{-\infty, 0} \cup \pco{1, \infty}, which is not open. For example, 11 is not in the interior of R∖E\R \setminus E.
  2. E=[0,1]E = \br{0, 1} is a closed set, but not open.
    • Its complement is R∖E=(−∞,0)∪(1,∞)\R \setminus E = \p{-\infty, 0} \cup \p{1, \infty}, which is open. However, EE itself is not open since, for example, 00 is not in the interior of EE.
  3. E=[0,1)E = \pco{0, 1} is neither open nor closed.
    • EE is not open since 00 is not in the interior of EE. On the other hand, its complement R∖E=(−∞,0)∪[1,∞)\R \setminus E = \p{-\infty, 0} \cup \pco{1, \infty} is not open, since 11 is not in the interior of X∖EX \setminus E.
  4. E=RE = \R is both open and closed.
    • EE is definitely open since every open ball is a subset of EE, i.e., a subset of the whole space. On the other hand, R∖R=∅\R \setminus \R = \emptyset, which is open. (This is an example of a vacuous truth. The empty set satisfies the definition of an open set because there are no elements to check the condition on.)

The next two propositions tell you how open/closed sets interact with unions/intersections:

Proposition

Let (X,d)\p{X, d} be a metric space.

  1. ∅,X\emptyset, X are both open.

  2. If {Ui}i∈I\set{U_i}_{i \in I} is a collection of open sets indexed by II, then its union ⋃i∈IUi\bigcup_{i \in I} U_i is also open.

    ‘‘Open sets are closed under arbitrary unions."``\text{Open sets are closed under arbitrary unions.}"

  3. If U1,…,UnU_1, \ldots, U_n are open sets, then their intersection U1∩⋯∩UnU_1 \cap \cdots \cap U_n is also open.

    ‘‘Open sets are closed under finite intersections."``\text{Open sets are closed under finite intersections.}"
Proposition

Let (X,d)\p{X, d} be a metric space.

  1. ∅,X\emptyset, X are both closed.

  2. If {Fi}i∈I\set{F_i}_{i \in I} is a collection of closed sets indexed by II, then its union ⋂i∈IFi\bigcap_{i \in I} F_i is also closed.

    ‘‘Closed sets are closed under arbitrary intersections."``\text{Closed sets are closed under arbitrary intersections.}"

  3. If F1,…,FnF_1, \ldots, F_n are closed sets, then their union F1∪⋯∪FnF_1 \cup \cdots \cup F_n is also closed.

    ‘‘Closed sets are closed under finite unions."``\text{Closed sets are closed under finite unions.}"

The takeaway from these propositions is this:

  1. Open sets interact nicely with unions.
  2. Closed sets interact nicely with intersections.

Adherent Points and Closure

Definition

Let (X,d)\p{X, d} be a metric space and E⊆XE \subseteq X be a subset. Then x∈Xx \in X (note that xx is not required to be an element of EE) is an adherent point of EE if for every r>0r > 0, we have B(x,r)∩E≠∅B\p{x, r} \cap E \neq \emptyset. The set of adherent points of EE is called the closure of EE and is denoted E‾\cl{E}.

Let's take another look at the definition of an adherent point:

for all r>0, B(x,r)∩E≠∅.\text{for all } r > 0,\ B\p{x, r} \cap E \neq \emptyset.

We can write it like this:

for all r>0, there exists y∈E such that d(x,y)<r.\text{for all } r > 0, \text{ there exists } y \in E \text{ such that } d\p{x, y} < r.

When you look at this definition, you should be thinking of what happens when rr is really small. Then you can interpret an adherent point as follows:

‘‘x is an adherent point of E if it can be approximated by elements in E."``x \text{ is an adherent point of } E \text{ if it can be approximated by elements in } E."

By quantifying what this means in different ways, you can find different ways of defining an adherent point:

Proposition

Let (X,d)\p{X, d} be a metric space and E⊆XE \subseteq X a subset. The following are equivalent:

  1. xx is an adherent point of EE.
  2. x∈E‾x \in \cl{E}.
  3. x∉Ext⁡Ex \notin \Ext{E}.
  4. There exists a sequence {xn}n\set{x_n}_n of elements in EE such that lim⁡n→∞xn=x\lim_{n\to\infty} x_n = x.
Remark.

To interpret (iii), if x∈Ext⁡Ex \in \Ext{E}, then xx is far away from EE, so it cannot be approximated by elements in EE. Thus, if x∉Ext⁡Ex \notin \Ext{E}, then it's close to EE, so it's an adherent point.

Examples

Example 3.

Let (X,d)=(R,∣ ⋅ ∣)\p{X, d} = \p{\R, \abs{\:\cdot\:}}. Show that Q‾=R\cl{\Q} = \R.
Solution.

Recall that Q\Q is dense in R\R: given any a<ba < b, there exists an element q∈Qq \in \Q such that a<q<ba < q < b. Rewriting,

a<q<b  ⟺  q∈(a,b),a < q < b \iff q \in \p{a, b},

so Q\Q intersects every open interval in R\R.

To show that Q‾=R\cl{\Q} = \R, we need to show that every element x∈Rx \in \R is an adherent point of Q\Q, i.e., for every r>0r > 0, we have B(x,r)∩Q≠∅B\p{x, r} \cap \Q \neq \emptyset. But B(x,r)=(x−r,x+r)B\p{x, r} = \p{x - r, x + r}, so by density of Q\Q, we see

B(x,r)∩Q=(x−r,x+r)∩Q≠∅,B\p{x, r} \cap \Q = \p{x - r, x + r} \cap \Q \neq \emptyset,

so x∈Q‾x \in \cl{\Q}. Thus, Q‾=R\cl{\Q} = \R, which was what we wanted to show.
Example 4.

(This example might be a little too hard for this class, but it's a good problem to chew on to test your analysis knowledge.)

Consider the metric space given by

X=ℓ1(N)={absolutely summable real sequences}={an ⁣:N→R | ∑n=1∞∣an∣<∞}d(a,b)=∑n=1∞∣an−bn∣.\begin{aligned} X &= \ell^1\p{\N} \\ &= \set{\text{absolutely summable real sequences}} \\ &= \set{\left. \func{a_n}{\N}{\R} \st \sum_{n=1}^\infty \abs{a_n} < \infty \right.} \\ d\p{a, b} &= \sum_{n=1}^\infty \abs{a_n - b_n}. \end{aligned}

Let

E={sequences in ℓ1(N) with at most finitely many non-zero terms}.E = \set{\text{sequences in } \ell^1\p{\N} \text{ with at most finitely many non-zero terms}}.

Show that EE is dense in ℓ1(N)\ell^1\p{\N}, i.e., show E‾=ℓ1(N)\cl{E} = \ell^1\p{\N}.
Solution.

We need to show, given a={an}n∈ℓ1(N)a = \set{a_n}_n \in \ell^1\p{\N} and ε>0\epsilon > 0, that

B(a,ε)∩E≠∅.B\p{a, \epsilon} \cap E \neq \emptyset.

Equivalently, we need to find an element b={bn}n∈Eb = \set{b_n}_n \in E (i.e., an element with at most finitely many non-zero terms) such that

d(a,b)<ε  ⟺  ∑n=1∞∣an−bn∣<ε.d\p{a, b} < \epsilon \iff \sum_{n=1}^\infty \abs{a_n - b_n} < \epsilon.

Let's work backwards to find bb: given any b∈Eb \in E, bb has only finitely many non-zero terms, so there exists NN such that bn=0b_n = 0 for n≥Nn \geq N. Thus,

d(a,b)=∑n=1∞∣an−bn∣=∑n=1N∣an−bn∣+∑n=N+1∞∣an−0∣=∑n=1N∣an−bn∣+∑n=N+1∞∣an∣.\begin{aligned} d\p{a, b} &= \sum_{n=1}^\infty \abs{a_n - b_n} \\ &= \sum_{n=1}^N \abs{a_n - b_n} + \sum_{n=N+1}^\infty \abs{a_n - 0} \\ &= \sum_{n=1}^N \abs{a_n - b_n} + \sum_{n=N+1}^\infty \abs{a_n}. \end{aligned}

To make this quantity small, remember that we have the freedom to choose what each bnb_n can be. So, we can simply set bn=anb_n = a_n for 1≤n≤N1 \leq n \leq N to make the first sum vanish:

d(a,b)=∑n=N+1∞∣an∣.d\p{a, b} = \sum_{n=N+1}^\infty \abs{a_n}.

To finish the proof, it suffices to show that there exists NN such that the tail of this sum is smaller than ε\epsilon. But the sequence aa is absolutely summable, which means that

lim⁡N→∞∑n=N+1∞∣an∣=0.\lim_{N\to\infty} \sum_{n=N+1}^\infty \abs{a_n} = 0.

Thus, we can find NN so that ∑n=N+1∞∣an∣<ε\sum_{n=N+1}^\infty \abs{a_n} < \epsilon.

Let's put everything together: let ε>0\epsilon > 0, and let NN be as in the previous sentence. Set

bn={anif 1≤n≤N,0if n>N.b_n = \begin{cases} a_n & \text{if } 1 \leq n \leq N, \\ 0 & \text{if } n > N. \end{cases}

By construction, b={bn}nb = \set{b_n}_n only has finitely many non-zero terms, so b∈Eb \in E. Moreover,

d(a,b)=∑n=1∞∣an−bn∣=∑n=N+1∞∣an∣<ε.\begin{aligned} d\p{a, b} &= \sum_{n=1}^\infty \abs{a_n - b_n} \\ &= \sum_{n=N+1}^\infty \abs{a_n} \\ &< \epsilon. \end{aligned}

So b∈B(a,ε)∩Eb \in B\p{a, \epsilon} \cap E, and because ε\epsilon was arbitrary, it follows that a∈E‾a \in \cl{E}. Since aa was also arbitrary, we conclude that E‾=ℓ1(N)\cl{E} = \ell^1\p{\N}.