Topology
Fernando Pazos Ruiz
24/2/2022
1 Accumulation points
Definition 1.1 (Accumulation point) \(\text{A point} \hspace{1.5mm} p \hspace{1.5mm} \text{of a metric space} \hspace{1.5mm} X \hspace{1.5mm} \text{is an accumulation point of} \hspace{1.5mm} E \subseteq X \hspace{1.5mm} \text{if} \hspace{1.5mm} \forall r > 0, \hspace{0.5mm} N_r(p) \cap E \neq \{p\} \hspace{1.5mm} \text{and} \hspace{1.5mm} \neq \emptyset.\)
That is, \(p\) is an accumulation point of some subset \(E\) of a metric space if within every neighborhood centered at \(p\) there is at least some element of \(E\) other than —in case \(p \in E\)— the point \(p\) itself.
In order for \(p\) to be an accumulation point of \(E\) it is not required, though, for \(p\) to actually belong to \(E\).
Proof. We will perform the proof of the contrapositive. Assume a radius \(r\) such that \(N_r(p)\) only contains a finite amount of points of \(E\) does exist. From this assumption we deduce the set \(N_r(p) \cap E\) is finite. Furthermore: if from this set we make sure to remove \(p\) —the center of the neighborhood— the resulting set will also be finite. That is, the set \(Q = \big(N_r(p) \cap E\big) \setminus \{p\}\) comprises a finite amount \(n\) of points —let’s denote them by \(q_1, \hspace{0.5mm} \ldots, \hspace{0.5mm} q_n\)—.
Consider now a neighborhood centered on \(\boldsymbol{p}\) whose radius is shorter than the distance between \(\boldsymbol{p}\) and the point \(\boldsymbol{q \in Q}\) closest to \(\boldsymbol{p}\). To make the argument clearer, consider the set of distances between \(p\) and each element of \(Q\). That set would be \(D = \{d(p, q_1), \hspace{0.5mm} \ldots, \hspace{0.5mm} d(p, q_n)\}\). Then, what we want is to think of a neighborhood around \(p\) whose radius is shorter than \(\boldsymbol{\min D}\). Consider, for instance, the radius \(h = \large \frac{\min D}{2}\).
The conclusion that we want to arrive to is that \(p\) is not an accumulation point of \(E\) —i.e. that within some neighborhood centered on \(p\) (not including the point \(p\) itself) there is no element of \(E\)—. To claim that \(\boldsymbol{p}\) is not an accumulation point of \(\boldsymbol{E}\) it suffices to provide the radius of one neighborhood whose only element shared with \(\boldsymbol{E}\) —if it even contains any element of \(\boldsymbol{E}\) at all— is its center \(\boldsymbol{p}\). Such a radius is \(\boldsymbol{h}\). We will argue right away why we deem \(\boldsymbol{h}\) to be the piece of evidence we require to sustain the claim that \(\boldsymbol{p}\) is not an accumulation point of \(\boldsymbol{E}\).
Since \(h < r\) it’s clear that \(N_h(p) \subset N_r(p)\). Consequently, the following holds.
\[\begin{equation} \large \overbrace{\boldsymbol{N_h(p) \cap E} = \big(N_h(p) \cap N_r(p)\big) \cap E}^{\text{Due to} \hspace{1mm} N_h(p) \hspace{0.5mm} \subset \hspace{0.5mm} N_r(p)}\underbrace{= \hspace{1mm} \boldsymbol{N_h(p) \cap \big(N_r(p) \cap E \big)}}_{\text{Due to the associativity} \\ \text{of the intersection of} \\ \text{sets}} \end{equation}\]
Now, there are two cases to be considered.
Case 1. \(\boldsymbol{p \notin E}\)
\(N_r(p) \cap E = Q\). Then, \(N_h(p) \cap E = N_h(p) \cap Q\), which in turn —since \(N_h(p)\) was crafted so that it would contain no element of \(Q\)— is the empty set.
Case 2. \(\boldsymbol{p \in E}\)
\(N_r(p) \cap E = Q \cup \{p\}\). Then, \(N_h(p) \cap E = \big(N_h(p) \cap Q \big) \cup \big(N_h(p) \cap \{p\}\big)\), which in turn is the union between the empty set and \(\{p\}\). Hence, it’s \(\{p\}\).
Corollary 1.1 (Only infinite sets can have any accumulation point)
\(\text{A finite subset of a metric space has no accumulation points.}\)
Definition 1.2 (Closed set) \(\text{The set} \hspace{1.5mm} \{p \in X \hspace{0.5mm} \big| \hspace{0.5mm} p \hspace{1.5mm} \text{is an accumulation point of} \hspace{1.5mm} E\} \hspace{1.5mm} \text{is usually denoted by} \hspace{1.5mm} E'.\)
\(\text{A subset} \hspace{1.5mm} E \hspace{1.5mm} \text{of a metric space} \hspace{1.5mm} X \hspace{1.5mm} \text{is closed if} \hspace{1.5mm} E' \subseteq E.\)
Though the definition is pretty much transparent, let’s just state it in everyday language for the sake of comprehension: \(E\) is closed if it contains all of its accumulation points.
Corollary 1.2 (Every finite subset is closed)
\(\text{If a subset of a metric space is finite then it is closed.}\)
Proof. Assume \(s\) is an accumulation point of \(E\). Then, for every \(r > 0\), \(\Big(N_r(s) \cap E \Big) \setminus \{s\} \neq \emptyset\). Particularly, for \(r = \large \frac{1}{n}\) this is satisfied: \(\Big(N_{\frac{1}{n}}(s) \cap E \Big) \setminus \{s\} \neq \emptyset\).
Let us define a function \(f: \mathbb{N} \rightarrow E, \hspace{1.5mm} f: n \mapsto s_n\) such that \(s_n \in \Big(N_{\frac{1}{n}}(s) \cap E \Big) \setminus \{s\}\).
Note that we are only able to define such a function because we are sure that, for every \(\boldsymbol{n \in \mathbb{N}}\), the set \(\boldsymbol{\Big(N_{\frac{1}{n}}(s) \cap E \Big) \setminus \{s\}}\) is not empty —otherwise, there would be no point \(\boldsymbol{s_n}\) in that set which \(f\) could return—.
On the other hand, for every \(\epsilon > 0\) this is true: \(\lceil \frac{1}{\epsilon} \rceil + 1 > \frac{1}{\epsilon}\). Then, the following is also true.
\[\begin{equation} \large \boldsymbol{\frac{1}{\lceil \frac{1}{\epsilon} \rceil + 1} < \epsilon} \end{equation}\]
With that in mind, let us craft for every \(\epsilon > 0\) a corresponding natural number \(N = \lceil \frac{1}{\epsilon} \rceil + 1\).
Regarding each \(s_n\), their membership to \(\Big(N_{\frac{1}{n}}(s) \cap E \Big) \setminus \{s\}\) implies \(s_n \in N_{\frac{1}{n}}(s)\) —and therefore \(d(s_n, s) < \frac{1}{n}\)—.
Consider \(n \geq N\). Then, the following holds.
\[\begin{equation} \large \overbrace{\boldsymbol{d(s_n, s)} < \frac{1}{n}}^{\text{Due to} \hspace{1mm} s_n \hspace{0.5mm} \in \hspace{0.5mm} N_{\frac{1}{n}}(s)} \underbrace{\leq \frac{1}{N}}_{\text{Due to} \\ n \geq N} \overbrace{< \boldsymbol{\epsilon}}^{\text{Due to} \\ \lceil \frac{1}{\epsilon} \rceil + 1 > \frac{1}{\epsilon}} \end{equation}\]
That is, for an arbitrary \(\epsilon > 0\), we have that if \(n \geq N\) then \(d(s_n, s) < \epsilon\). And with that we conclude the proof.
2 Interior points
Definition 2.1 (Interior point) \(\text{A point} \hspace{1.5mm} p \hspace{1.5mm} \text{of a subset} \hspace{1.5mm} E \hspace{1.5mm} \text{of a metric space} \hspace{1.5mm} X \hspace{1.5mm} \text{is an interior point of} \hspace{1.5mm} E \hspace{1.5mm} \text{if} \hspace{1.5mm} \exists r > 0 \hspace{1.5mm} \text{such that} \hspace{1.5mm} N_r(p) \subseteq E.\)
In plain words: for \(p\) to be an interior point of \(E\) at least one of the neighborhoods around \(p\) must be a subset of \(E\).
Exercise 2.1 \(\text{Consider the notation int} \hspace{0.5mm} E \hspace{1.5mm} \text{to refer ourselves to the set of interior points of} \hspace{1.5mm} E \hspace{1.5mm} \text{—also known as the interior of} \hspace{1.5mm} E \text{—.}\)
\(\text{Prove that} \hspace{1.5mm} \text{int} \hspace{0.5mm} (\text{int} \hspace{0.5mm} E) = \hspace{0.5mm} \text{int} \hspace{0.5mm} E.\)
Solution. The idea is to prove both \(\text{int} \hspace{0.5mm} (\text{int} \hspace{0.5mm} E) \subset \text{int} \hspace{0.5mm} E\) and \(\text{int} \hspace{0.5mm} E \subset \text{int} \hspace{0.5mm} (\text{int} \hspace{0.5mm} E)\).
Proving \(\boldsymbol{\text{int} \hspace{0.5mm} (\text{int} \hspace{0.5mm} E) \subset \text{int} \hspace{0.5mm} E}\)
Assume \(p\) is an interior point of the set \(\text{int} \hspace{0.5mm} E\). That is, there exists a radius \(r\) such that \(N_r(p) \subseteq \hspace{0.5mm} \text{int} \hspace{0.5mm} E\). Therefore, every point belonging to the neighborhood —including the center \(p\) itself— belongs to \(\text{int} \hspace{0.5mm} E\). And with that we conclude that membership of an arbitrary point \(p\) to \(\text{int} \hspace{0.5mm} (\text{int} \hspace{0.5mm} E)\) implies membership of that very same point to \(\text{int} \hspace{0.5mm} E\).
Proving \(\boldsymbol{\text{int} \hspace{0.5mm} E \subset \text{int} \hspace{0.5mm} (\text{int} \hspace{0.5mm} E)}\)
Let us express the proposition in these other terms: \(p \in \text{int} \hspace{0.5mm} E \Rightarrow p \in \text{int} \hspace{0.5mm} (\text{int} \hspace{0.5mm} E)\). We will prove the contrapositive instead: \(p \notin \text{int} \hspace{0.5mm} (\text{int} \hspace{0.5mm} E) \Rightarrow p \notin \text{int} \hspace{0.5mm} E\).
Assume \(p\) is not an interior point of the set \(\text{int} \hspace{0.5mm} E\). That is, for every radius \(r > 0\) the neighborhood \(N_r(p)\) is not a subset of \(\text{int} \hspace{0.5mm} E\). In other words, \(N_r(p) \setminus \hspace{0.5mm} \text{int} \hspace{0.5mm} E \neq \emptyset\). Another way of conveying the same idea: for an arbitrarily narrow neighborhood \(\boldsymbol{N_r(p)}\) there exists a point \(\boldsymbol{q \in N_r(p)}\) such that \(\boldsymbol{q \notin \hspace{0.5mm} \text{int} \hspace{0.5mm} E}\).
To say that within a neighborhood centered on \(p\) which displays an arbitrarily small radius \(r\) there exists a point \(q\) is to say that \(q\) satisfies \(\forall r > 0, \hspace{1mm} d(q, p) < r\).
There is a useful lemma that we have proven in a previous notebook —the one dealing with convergence—. It will come in handy now.
\[\begin{equation} \text{Any two numbers} \hspace{1.5mm} x \hspace{1.5mm} \text{and} \hspace{1.5mm} y \hspace{1.5mm} \text{satisfy} \hspace{1.5mm} x = y \hspace{1.5mm} \text{if and only if} \hspace{1.5mm} \forall \epsilon > 0, \hspace{0.5mm} |x - y| < \epsilon. \end{equation}\]
Due to this lemma we conclude \(q = p\). Hence, what was said about \(q\) —namely, that it does not belong to \(\text{int} \hspace{0.5mm} E\)— also applies for \(p\). And with that we have accomplished the proof.
Example 2.1 \(\text{The interior of any subset} \hspace{1.5mm} E \hspace{1.5mm} \text{of a metric space} \hspace{1.5mm} X \hspace{1.5mm} \text{is an open set.}\)
The statement above was in actuality verified on Exercise 2.1. Since \(\text{int} \hspace{0.5mm} (\text{int} \hspace{0.5mm} E) = \hspace{0.5mm} \text{int} \hspace{0.5mm} E\), every point belonging to the set \(\text{int} \hspace{0.5mm} E\) is an interior point of that very same set.
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