Lecture 21 - More Definitions on Topology

We defined:

Isolated Point

A point $a \in A$ is an isolated point of $A$ if it is not a Limit Point of $A$ . Namely, $\exists ε > 0 (V_{ε} (a) \cap A = \emptyset \lor {x})$ .

The definition here is to avoid the case in the limit pointer where we have the sequence $(a, a, a, \dots)$ , since that is technically a limit point.

You can think of having some space between $a$ and all the contiguous points of $A$ .

It's literally isolated! For more definitions:

Closed (Toplogical)

A set $F \subset R$ is closed if it contains its Limit Points.

Closure

$A \subset R$ The closure of $A$ is:
$\overset{―}{A} = A \cup {x | x is a limit point of A}$

We then presented the following theorem:

Closure implies Cauchy Sequences

$A$ is Closed (Toplogical) ifff every Cauchy Sequence in $A$ converges to a point in $A$ .

Proof

See 32 Open and Closed Sets Practice#3.2.5:

3.2.5

Question

Prove Closed iff Convergent Cauchy Sequences.

Closure implies Cauchy Sequences

$A$ is Closed (Toplogical) ifff every Cauchy Sequence in $A$ converges to a point in $A$ .

Proof

( $\Rightarrow$ ): Suppose $A$ is closed, so then $A$ contains it's limit points. Let $(x_{n}) \subseteq A$ be a Cauchy Sequence. Then $(x_{n})$ converges by being Cauchy. Say it converges to $x$ , then since $A$ is closed then $x \in A$ as required since $A$ contains its limit points, which are equivalent to the limits of the convergent sequences of $A$ .

( $\Leftarrow$ ): Suppose $\forall (x_{n}) \subseteq A$ where $(x_{n}) \to x$ (ie: the sequence $(x_{n})$ is Cauchy because it converges) where $x \in A$ . We need to show that $A$ contains it's limit points.

Let $ℓ$ be a limit point of $A$ . We want to show that $ℓ \in A$ . Since $ℓ$ is a limit point of $A$ then equivalently there is some sequence of points in $A - {ℓ}$ where it converges to $ℓ$ . Denote this sequence $(a_{n}) \in A - {ℓ}$ where $(a_{n}) \to ℓ$ . then since all convergent sequences in $A$ converge to a point in $A$ then $ℓ \in A$ .

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See some examples at Lecture 21 - More Definitions on Topology#Examples.

This implies that, for example, The Density of Q in R can be reformulated as:

(Density of

Q

R

, via Sequences)

For every $y \in R$ then $\exists (x_{n}) \subseteq Q$ as a Sequence such that it converges to $y$ .

This suggests that we can always find some rational approximation for a real number, no matter what!

Examples

Let $a < b$ . Then

$[a, b]$ is closed.
For $x \in R$ , then ${x}$ is closed (the limit points are only $x$ which is in the set).
$R$ is definitely closed (any limit point is a real number)

Theorem

Let $A \subseteq R$ . Then $\overset{―}{A}$ (the closure of $A$ ) is the smallest closed set containing $A$ .

The proof isn't super trivial, since we might accidentally throw in too many points.

Proof

If $L$ is the set of limit points of $A$ , then it is immediately clear that $\overset{―}{A}$ contains the limit points of $A$ . There is still something more to prove, that because taking the union of $L$ with $A$ then we could potentially produce some new limit points of $\overset{―}{A}$ . See 32 Open and Closed Sets Practice#3.2.7 for practice on this.

Now any closed set containing $A$ must contain $L$ as well. This shows that $\overset{―}{A} = A \cup L$ is the smallest closed set containing $L$ .

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Some Notes on Open/Closed

While gramatically we may think open/closed here implies they are opposites, that isn't the case. There are cases where a set may be neither open nor closed such as:

(a, b]

Isn't closed since $a$ is a limit point and isn't in the set
Isn't open via our reasoning in Open Sets

Another example would be:

{\frac{1}{n} | n \in N}

Isn't closed since $0$ is a limit point and isn't in the set
Isn't open since there are points like $\frac{1}{2}$ that are in an $V_{ϵ} (\frac{1}{2})$ with no elements from the set intersected with it.

Some sets can be both open and closed. Some examples are:

$R$
$\emptyset$ (there are no limit points so it's closed, and it's open since there are no elements to satisfy the definition).

In the HW we show that these are the only two sets that are subsets of $R$ that do this.

Complement

A^{c}

If $A \subseteq R$ then:
$A^{c} = R - A$

See Closed Implies Complement is Open (Vice Versa) for an application of this.

Theorem

Let $A \subseteq R$ . Then $A$ is open iff $A^{c}$ (the Complement) is Closed (Toplogical). This is the same as saying that $A$ is closed iff $A^{c}$ is open.

Proof

( $\Rightarrow$ ): Suppose $A$ is open. Let's show $A^{c}$ is closed.

Let $x$ be a limit point of $A^{c}$ . We want to show that $x \in A^{c}$ , so assume for contradiction that $x \in A$ , so then $\exists ϵ > 0$ where $V_{ϵ} (x) \subseteq A$ . But then for any sequence of points in $A^{c}$ , no element can be within $ϵ$ distance of $A$ , so $x$ cannot be a limit point of $A$ . This is a contradiction, so $x \notin A$ .

( $\Leftarrow$ ): Suppose $A^{c}$ is closed. Let $y \in A$ . Then $y$ is not a limit point of $A^{c}$ . Then $\exists ϵ > 0$ such that:
$V_{ϵ} (y) \cap A^{c} = \emptyset$
But if that's the case then $V_{ϵ} (y) \subseteq A$ as desired.

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Theorem

An arbitrary insersection of Closed (Toplogical) sets is closed.
A finite union of closed sets is closed.

Proof

Apply DeMorgan's Law with Closed Implies Complement is Open (Vice Versa) and use an induction. See also Union & Intersection of Collections of Open Sets for a similar proof.

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compact set, compactness

Let $K \subseteq R$ . $K$ is compact if every Sequence of points in $K$ has a subsequence which converges to a point $k \in K$ .

Some examples include:

Let $a < b$ . Then $[a, b]$ is compact. Why? If I have a sequence of points $(x_{n}) \in [a, b]$ for all $n \in N$ then $\forall n \in N$ then $a \leq x_{n} \leq b$ . This means the sequence is bounded. So then apply the Balzano-Weierstrass Theorem to get that there $\exists$ a convergent subsequence $(x_{n_{k}})$ that converges.

But for each $k \in N$ then $a \leq x_{n_{k}} \leq b$ still. We know $(x_{n_{k}})$ is convergent, so say $(x_{n_{k}}) \to x$ so by the Limits and Order (Order Limit Theorem) then $a \leq x \leq b$ . We know $x \in [a, b]$ as a result.
As a counter example $R$ is not compact. The sequence of numbers:
$(1, 2, 3, 4, \dots) \subseteq R$
has no convergent subsequence.

Nested Compact Set Property

If:
$K_{1} \supseteq K_{2} \supseteq \dots$
are nested, compact, nonempty sets. Then:
$⋂_{n = 1}^{\infty} K_{n} \neq \emptyset$

Sketch of the proof:

Notice $\forall n \in N$ then let $x_{n} \in K_{n}$ . Note $x_{1} \in K_{1}$ , $x_{2} \in K_{2} \Rightarrow K_{1}$ , $x_{3} \in K_{3} \Rightarrow x_{3} \in K_{2} \Rightarrow x_{3} \in K_{1}$ , and so on. Then each $x_{n} \in K_{1}$ . Therefore, then $\exists$ a convergent subsequence $(x_{n_{i}})$ where $x = lim_{i \to \infty} x_{n_{i}} \in K_{1}$

Now $x_{n_{1}} \in K_{n_{1}} \subseteq K_{1}$ since $n_{1} \geq 1$ . Similarly $x_{n_{2}} \in K_{n_{2}} \subseteq K_{2}$ since $n_{2} \geq 2$ . In general $x_{n_{i}} \in K_{i}$ for all $i$ since $n_{i} \geq i$ . Now:
$x = lim_{i \to \infty} x_{n_{i}} = (x_{n_{2}}, x_{n_{3}}, \dots) \in K_{2}$

We can repeat this process, where the next step shows:
$x = lim_{i \to \infty} x_{n_{i}} \in K_{i}$
so $x \in K_{i}$ for all $i$ so then their intersection is non-empty.

Proof

For each $n \in N$ pick a point $x_{n} \in K_{n}$ . Because the compact sets are nested then $(x_{n}) \subseteq K_{1}$ and by the definition of compactness then there is a convergent subsequence $(x_{n_{k}})$ whose limit $x \in K_{1}$ .

But you can repeat this process for each $K_{n}$ . To show this use a particular $n_{0} \in N$ so the terms in the sequence $(x_{n}) \subseteq K_{n_{0}}$ when $n \geq n_{0}$ , ignoring the finite number of terms at the start. Thus then the same subsequence $(x_{n_{k}}) \subseteq K_{n_{0}}$ , so then $x$ is a limit of each $(x_{n_{k}}) \subseteq K_{n_{0}}$ . Since $n_{0}$ was arbitrary then $x \in ⋂_{n = 1}^{\infty} K_{n}$ .

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This idea is similar to the Nested Interval Property, except it works with generic sets.