Lecture 20 - Finishing Cantor Set, Review of Ch. 2

Recall from last time

Recall the definition of the Epsilon-Neighborhood of $x$ :

ε

-neighborhood of

a

$V_{ε} (a) = (a - ε, a + ε) = {x \in R | | x - a | < ε}$

open set

Let $U \subseteq R$ . Then $U$ is open if:
$\forall x \in U \exists ε > 0 (V_{ε} (x) \subseteq U)$
ie: each $x \in U$ has some $ε$ -neighborhood contained in $U$ .

Example: $(a, b)$

$(a, b)$ is open.

Since choose $ε = min (| x - a |, | x - b |)$ implies that $V_{ε} (x) \subseteq (a, b)$ .

By constrast if we considered even a one-sided interval like $(a, b]$ then it is NOT open:

Notice if $x = b$ then any $\varepsilon > 0$ would have some element in $V_\varepsilon(b)$ be outside the interval.

To elaborate on the limit point, it's saying that $\forall ϵ > 0$ :

(V_{ϵ} (x) ∖ {x}) \cap A \neq \emptyset

For instance, we showed that $(a, b]$ is open, and that the set of limit points of $A$ is equal to $[a, b]$ . This brings us to a new theorem:

Alternative Definition of Limit Point

Let $A \subset R$ and $x \in R$ . Then $x$ is a limit point of $A$ iff $\exists (a_{n})$ :
$a_{n} \in A ∖ {x}$
for all $n$ and $a_{n} \to x$ .

Proof

( $\Rightarrow$ ): Suppose $x$ is a limit point of $A$ . Let's apply a sequence of neighborhoods such that we can construct $a_{n}$ . To do this, let $n \in N$ . Then choose:
$a_{n} \in (V_{\frac{1}{n}} (x) ∖ {x}) \cap A$
we claim that $a_{n} \to x$ . To prove this via the definition let $ϵ > 0$ . Choose $N \in N$ such that $\frac{1}{N} < ϵ$ . Then suppose $n \geq N$ then:
$\begin{aligned} | a_{n} - x | & < \frac{1}{n} & (a_{n} \in V_{\frac{1}{n}} (x)) \\ \leq \frac{1}{N} \\ < ϵ \end{aligned}$
( $\Leftarrow$ ): Suppose $\exists a_{n}$ where $a_{n} \in A - {x}$ for all $n$ and that $a_{n} \to x$ . Let $ϵ > 0$ so that $V_{ϵ} (x)$ is an epsilon-neighborhood of $x$ . We want to show that $(V_{ϵ} (x) - {x}) \cap A \neq \emptyset$ . We can choose a $N \in N$ such that:
$| a_{n} - x | < ϵ$
by the definition and that $a_{n} \to x$ . Then definitely $| a_{N} - x | < ϵ$ , so then $N \in V_{ϵ} (x)$ . Furthermore clearly $a_{N} \in A$ while $a_{N} \neq x$ so then $a_{N} \in (V_{ϵ} (x) - {x}) \cap A$ is not empty as desired.

☐

To illustrate this, consider:
$A = {\frac{1}{n} | n \in N}$
Then the set of all limit points of $A$ is just ${0}$ . Notice other points like $\frac{1}{2}$ are not in this set since there's no sequence of numbers constructing from the set cannot approach $\frac{1}{2}$ .

Another example is when $A = Q$ . The set of limit points of $A$ are now $R$ .

Review of Ch. 2

We talked about the following problems:

Example: (a,b)

Alternative Definition of Limit Point

Review of Ch. 2

Example: $(a, b)$