Proof tips

• $R - f^{-1}(R - S) = f^{-1}(S)$. You can show this by contradiction

Why topology?

We can provide good definitions of continuity for $R$, but when it comes to arbitrary spaces like $R^2$ or funky regions, we are at a loss. That’s where topology comes in!

Set interiors

Definition of interiority 🐧

The interior of a set $S$ is the set of all $x \in S$ such that there exists some $\epsilon > 0$ where $(x - \epsilon, x + \epsilon) \in S$. This () notation denotes an open interval. We denote this as $int(S)$.

Implications of interiors 🚀

Let’s say that $S = [0, 1]$. The $int(S)$ you can construct $\epsilon = \min(x, 1-x)$. Now, this is valid for all $x \in S$ EXCEPT for $0, 1$, where the $\epsilon = 0$, which violates the rules. So, you say that $int(S) = (0, 1)$. s

Let’s say that $S = (0, 1)$. Now, you can construct the same $\epsilon$, and that always yields the interval $(x - \epsilon, x + \epsilon) = (0, 1) \in S$, which means that $int(S) = S = (0, 1)$.

So, the conclusion is that open sets are its own interior! (more on this later)

• Complicated example

  Say that we have $S = [-1, 1] \backslash \{1/n, n \in \mathbb{N}\}$. This is the uncountable interval with a countably infinite elements removed from it. We want to show that $0 \notin int(S)$.

  We can go back to the original definitions and say for contradiction that there does exist some $\epsilon$ such that $(0 - \epsilon, 0 + \epsilon) \in S$. However, we know that for all $\epsilon$, we have $|1/n| < \epsilon$ for some $n > N_\epsilon$. Therefore, the interval $(-\epsilon, \epsilon)$ will never purely contain elements from $S$. In other words, the interval will always contain some $1/n$. Therefore, we conclude that $0$ is not in the interior of $S$.

Limit Points

Definition of limit points 🐧

Given that $S \subseteq \mathbb{R}$, then a limiting point of $S$ is some $x$ where for any $\epsilon > 0$, there exists some $s \in S$ where

• Example: $S = \{1/n, n \in N\}$

  The limiting point of $S$ is $0$. This is because you have $|1/n - 0| < \epsilon$, and this is actually just the limit definition! We can easily set $n > \frac{1}{\epsilon}$, and this satisfies the inequality.

  In some ways, it’s a looser (existential) way of defining the limit.

We need to use the $<$ and not $\leq$, because if $s = x$, we get these things called isolated points. In the example above, $1$ would be an isolated point, as for any arbitrary $\epsilon > 0$, we can just select $s = 1/1$, and the absolute value is zero. Instead, we wnat to focus on points that have a place of contraction, where things squeeze towards it, rather than going away.

Finding limit points 🔨

A point $x$ is a limiting point if you can find a sequence of points in $S$ such that $s_n \neq s, \lim s_n = x$. The proof is literally the definition of the limit points and the limit definition $|s_n - x| < \epsilon$.

Set Closures 🐧

We can create a set closure by doing $\bar S = S \cup \{LP(S)\}$, where $LP$ are the limiting points of $S$.

• Examples of how we can close a set

  Continuing with the $1/n$ set above, the closed set would just $\bar S = \{1/n, n \in N\} \cup \{0\}$.

  If we have $S = (0, 1)$, then $0$ is a limiting point because you can always find $|0 - \frac{1}{n}| < \epsilon$ by setting $n = 1/\epsilon$. Similarly, you can do it with $1$, so the closed set is $[0, 1]$. Or, using the definition we just created, we can just let $s_n = 1/n$ and you are done.

  Actually, in the above example, anything in $(0, 1)$ is also a limit point, but this is already covered by $S$, so we ignored it. The only new thing that isn’t covered by $S$ is the 0 and 1.

It follows that an already closed set will have all $LP(S) \in S$.

• Examples of closed sets

  $\{1\}$ is closed, because you can’t find any such sequence

  $[0, 1]$ is closed. Any point in $[0, 1]$ is a limiting point, but the union of the sets stay the same.

Open and Closed Intervals

Finding open intervals 🔨

We say that a set is open if $S = int(S)$. Remember that we have a specific definition for the interiority of a set.

• $S = \mathbb{R}, S = \emptyset$ is open

• Any union of open sets is also open

Finding closed intervals 🔨

To show closure, you can do the following (equivalent)

• Show that $S = \bar S$, i.e. the closure of $S$ is the same as $S$

• Show that $\mathbb{R} - S$ is open
  • Proof

    We just want to show that for all $x \in R - S$, there exists some $\epsilon$ such that $(x - \epsilon, x + \epsilon) \in R - S$.

    Suppose for contradiction that this is not true, i.e. there exists some $x$ such that $(x - \epsilon, x + \epsilon) \notin R - S$ for all $\epsilon$. Now, let $\epsilon = 1/n$. This means that in every $(x - 1/n, x + 1/n)$ we can find some $s_n \in S$ in this interval.

    By the sandwich theorem, we know that $\lim s_n = x$, which means that $x$ is a limiting point in $S$. Because $S$ is closed, then it contains all limiting points, which means that $x \in S$. This is a contradiction, as we previously assumed that $x \in R - S$.

    So, we must conclude that if $S$ is a closed set, then $R - S$ is an open set.

• Show that any convergent sequence in $S$ has a limit in $S$

Some critical results 🚀

Union of any open intervals is open.

• Proof

  Pick any $x$ in the union. This $x$ must belong to one open interval. Therefore, it is on the interior of that interval, and so it’s on the interior of the interval union, and we are done.

intersection of finite open intervals is open

• Proof

  For every $x$ in the intersection, we have $x \in A_1, …, A_n$. There must exist one $\epsilon_1, … \epsilon_n$ for each set such that $(x - \epsilon_1, x + \epsilon_1) \in A_1$, and so on and so forth. Pick $\min(\epsilon_1, …, \epsilon_n)$. This logic breaks down for infinite intersections.

intersection of any closed intervals is closed

• Proof (using open intervals and complements)

  The complement of the intersection of closed intervals is equivalent to the union of these intervals’ complements, i.e.

  $$R - \bigcap A_n = \bigcup (R - A_n)$$

  Any union of open intervals are open, so $R - \cap A_n$ is open. Therefore, $\cap A_n$ is closed.

Union of finite closed intervals is closed

• Proof (using open intervals and complements)

  The complement of the the union of closed intervals is the intersection of the complements of closed intervals, which we know are open. The finite intersection of open intervals is open. So, the complement of the union of closed intervals is open, so the union of closed intervals is closed.

Why isn’t the union of infinite closed intervals closed? Well, you can think of any set as the infinite union of individual elements, and not all sets are closed!

Compact Sets and Open Covers

Compact Set Definition 🐧

There is a less formal definition that we use in $R$, and then after we talk about open covers, we will propose a more formal definition that works in any metric space.

The less formal definition (works in $R$ only): a compact set is such that if we have an arbitrary sequence $s_n$ from $S$, we can find a convergent subsequence whose limit is in $S$. We will mostly be using this definition in class.

• Examples (easier to show that something is not a compact set)

  The interval $(0, 1)$ is not compact because you can find $s_n = 1/n$, and $\lim s_n = 0 \notin (0, 1)$.

  The interval $\{x \geq 0, x \in \mathbb{R}\}$ is not compact because you can let $s_n = n$, and $\lim s_n = \infty \notin \mathbb{R}$.

The formal definition is as follows: A compact set is such that every open cover of $S$ has a finite subcover

• Examples

  The closed interval $[-100, 100]$ we can start with $U_\alpha = (-n, n)$ and take the subset $I’ = \{x \leq 100, x \in \mathbb{N}\}$. It’s not completely trivial to show that all open covers have a finite subcover

  The open interval $(0, 1)$ isn’t compact because we can have $U_\alpha = (1/n, 1)$ such that we need $I = N$ to construct an open cover of $(0, 1)$ and no subcover can work.

  Again, these just scratch the surface. Proving compactness is not trivial!

Heine-Borel Theorem 🚀

$S$ is compact if and only if $S$ is closed and bounded

• Partial proof

  We will only prove one special case in the backward direction (closed and bounded → compact) and only for $R$, but this theorem applies to all metric spaces.

  Say that we have a closed and bounded set $[a, b]$. Without loss of generality, we can set $a = 0, b = 1$.

  Let’s start with an arbitrary sequence $s_n \in [0, 1]$. By defiition, if we split the interval up into $[0, 1/2]$ and $[1/2, 1]$, at least one of these intervals must have an infinite number of terms (otherwise, the whole interval would only have a finite number of terms)

  Take the infinite interval, split it again, and the same logic applies. You can keep on splitting and choosing the infinite portion, and you will arrive at a single number, which is the limit. This limit value must be in $S$, by construction. This is the definition of a compact set, so we are done.

Open Covers 🐧

An open cover of a set $S$ is defined as a collection of open sets whose union contains $S$, or in symbolic terms:

$U = \{U_\alpha | \alpha \in I\}$ such that $S \subseteq \bigcup_{\alpha \in I} U_\alpha$.

The $I$ can be a finite set, or it can be something like $N$ or $R$.

• Examples

  An open cover of $R$ we can consruct as $U_n = (-n, n)$ and $I = N$.

  Another open cover of $R$ we can construct as $U_n = (n - 1, n + 1)$ and $I = R$.

A subcover is just a subset $I’$ and we take the same union but across this $I’ \subseteq I$.

Metric Spaces (bonus)

What is a metric space? 🐧

Let’s try to generalize some things to $R^n$. In $R$, there was a natural ordering, and this was important to many of our proofs (think: triangle inequality). But as we move to arbitrary sets like $R^n$ or even some wacky one like an arbitrary $S$, this ordering doesn’t exist anymore.

Instead, we have to define it through a distance function. A distance function, or metric, has these properties

1. $d(x, x) = 0$

2. $d(x, y) = d(y, x)$

3. $d(x, z) \leq d(x, y) + d(y, z)$ (triangle inequality)

• Examples of metric spaces

  We know that $R^n$ is a metric space. We can define several metrics on it, including the vector norm. The triangle inequality holds here too, although it is not entirely obvious.

We call a metric space as complete if every Cauchy sequence in $S$ converges to some element in $S$. The metric space $(R, abs)$ is complete because we have shown that cauchy sequences implies convergence.

Sequences in metric spaces 🚀

A sequence in any metric space converges to $s$ if $\lim d(s_n, s) = 0$. You can set up all the traditional limit definitions. If the metric space is complete, you can also use cauchy definitions.

For $R^n$, you can actually show that a sequence converges if its individual dimensions are convergent. Through this, you can show that $R^n$ is complete

The Bolzano-Weierstrass Theorem also holds in $R^n$. All of these proofs in $R^n$ hinges on the fact that you can separate things by dimensions.

Interiority in metric spaces 🚀

this is actually not too hard. Just replace all our discussions of $(x -\epsilon, x + \epsilon)$ with the notion of $d(x, x_0) < \epsilon$.