A summary of Chapter 3 of the book “Topology without Tears” by Sidney Morris that I’m reading.

## Limit Points

Definition. Let A be subset of a topo space $(X, \mathcal{T})$. A point $x\in X$ is called a limit point of A (aka accumulation point or cluster point) if $\forall U\in\mathcal{T}$ that contains x contain a point of A other than x. In other words,

$(U\cap A)-\{x\}\neq\emptyset\Leftrightarrow(U-\{x\})\cap A\neq\emptyset$

Theorem. $A\subseteq X$ is a closed set in the topo space $(X,\mathcal{T})$ iff A contains all of its limit points. ▶ Proof

Suppose that A is closed and $x\notin A$ but $x$ is a limit point of A. Therefore, $\forall U\in\mathcal{T}$ that contains x contain a point of A other than x. However, $X-A$ is an open set containing x but does not intersect A. Hence x is not a limit point, which contradicts to the assumption. Hence if A is closed then it contains all of its limit points.

Suppose that A contains all of its limit points. For any $x\notin A$, x is not a limit point, so $\exist U_x\in\mathcal{T}$ such that $x\in U_x$ and $U_x\cap A=\emptyset$, i.e. $U_x\subseteq X-A$. We can write X-A as an infinite union $\bigcup_{x\notin A}U_x$ of open sets, hence X-A is open and A is closed.

Note that if $x\in A$ doesn’t guarantee it is a limit point. For example, if A is a finite subset of the topology $\mathbb{R}$, no point in A is a limit point, neither is any other point in $\mathbb{R}$ is a limit point. Because for every $x\in\mathbb{R}$, you can always find an interval $(a,b)$ containing x but intersect $A-\{x\}$ trivially.

Another example is that in the discrete topology $(X,\mathcal{T})$, no point is a limit point of any set. Recall that the basis of the discrete topology is the set of singleton sets. For any point x and a subset A of X, $\{x\}$ is an open set containing x but $(A\cap\{x\})-\{x\} = \emptyset$.

## Closure

Theorem. Let A’ be the set of all limit points of A in the topo space $(X,\mathcal{T})$. Then $\bar{A}=A\cup A'$ is a closed set and it is called the closure of A.

This gives us some conclusions:

•  $\bar{A}$ is the smallest closed set containing A.
• Every closed set containing A must also contain $\bar{A}$.
•  $\bar{A}$ is the intersection of all closed sets containing A.
• A is closed iff A is a closure of itself.

Examples:

• The closure of $A=\left\{1,\frac12,\frac13,\dots,\frac1n,\dots\right\}$ in $\mathbb{R}$ is ▶ Reveal

It has been proven that $A$ is not a closed set, but $A\cup\{0\}$ is a closed set, and also the smallest closed set containing A, hence it is the closure of A.

• The closure of $\mathbb{Q}$ in $\mathbb{R}$ is $\mathbb{R}$. ▶ Proof

Suppose that $\bar{\mathbb{Q}}\neq\mathbb{R}$, i.e. $\forall x\in \mathbb{R}-\bar{\mathbb{Q}}$ (an open set) we have $a<b,\ a,b\in\mathbb{R}$ such that $x\in (a,b)\subseteq \mathbb{R}-\bar{\mathbb{Q}}$. However this interval also contains rational numbers, which contradicts to $(a,b)\subseteq\mathbb{R}-\bar{\mathbb{Q}}$. Hence $\bar{\mathbb{Q}} = \mathbb{R}$.

• The closure of $\mathbb{Z}$ in $\mathbb{R}$ is $\mathbb{Z}$ because it is a closed set.
• The closure of $\mathbb{I}$ in $\mathbb{R}$ is $\mathbb{R}$ because any interval $(a,b)$ contains infinitely many irrational numbers, hence intersect $\mathbb{I}$ non-trivially.

Definition. Let A be a subset of a topo space $(X,\mathcal{T})$. We say A is dense in X (aka everywhere dense) if $\bar{A}=X$.

Theorem. A is dense iff every non-empty open set intersects A non-trivially. ▶ Proof

Suppose that A is dense, i.e. $\bar{A}=X$, but there exists a non-empty open set U that $U\cap A=\emptyset$, so $x\in U$ is not a limit point of A. This contradicts to the assumption that every point in X is a limit point. Hence every non-empty open set intersects A non-trivially.

Suppose that every non-empty open set intersects A non-trivially. For any point $x\in X-A$, any open set U containing x is surely non-empty, so $U\cap A\neq\emptyset$. Since $x\notin A$, we have $U\cap A-\{x\}\neq\emptyset$, hence x is a limit point. Now we have that all points in X-A are limit points of A. The closure of A is hence $\bar{A}=(X-A)\cup A=X$.

Theorem. $\overline{A\cap B}\subseteq\overline{A}\cap\overline{B}$. ▶ Proof

Let $x$ be a limit point of $A\cap B$. We have that $\forall S$ open and containing x:

$(S-\{x\})\cap A\cap B\neq\emptyset\\ \Leftrightarrow \left\{\begin{matrix} (S-\{x\})\cap A\neq\emptyset\\ (S-\{x\})\cap B\neq\emptyset \end{matrix}\right.$

That means x is a limit point of A and a limit point of B. $\Rightarrow x\in\overline{A}\cap\overline{B}$.

Now we shall prove that $A\cap B\subseteq\overline{A}\cap\overline{B}$.

$\overline{A}\cap\overline{B}=(A\cup A')\cap\overline{B}\\ = (A\cap\overline{B})\cup(A'\cap\overline{B})\supseteq A\cap\overline{B}\\ = A\cap(B\cup B')=(A\cap B)\cup (A\cap B')\\ \supseteq (A\cap B)$

The theorem has been proven.

An example of $\overline{A\cap B}\neq\overline{A}\cap\overline{B}$ is when $A=\mathbb{I}$ and $B=\mathbb{Q}$. The LHS is $\emptyset$ while the RHS is $\mathbb{R}$.

Theorem. Let S be a dense subset of a topo space $(X,\mathcal{T})$. For every open set $U$, $\overline{S\cap U}=\overline{U}$. ▶ Proof

The proposition is obviously true when $U = \emptyset$, hence we only take care of the case $U\neq\emptyset$ below.

First, $\overline{S\cap U}\subseteq\overline{S}\cap\overline{U}=X\cap\overline{U}=\overline{U}$

We need to prove $\overline{U}\subseteq\overline{S\cap U}$.

Let $x\in \overline{U}$ be a limit point of $U$. We need to prove that $x\in\overline{S\cap U}$, i.e. $x$ is a limit point of $S\cap U$ or belongs to $S\cap U$. Anyway, we have that for any open set A containing x, $(A-\{x\})\cap U\neq\emptyset$.

• If $x$ is a limit point of $S$, we also have $(A-\{x\})\cap S\neq\emptyset$. Moreover, since $S$ is dense and $U$ is open, we have $S\cap U\neq\emptyset$. This implies that $(A-\{x\})\cap S\cap U\neq\emptyset$. Hence $x$ is a limit point of $S\cap U$. $\Rightarrow x\in\overline{S\cap U}$.
• If $x$ is not a limit point of $S$, for sure we have $x\in S$ (because S is dense, if x was not in S then x would be a limit point). That means there exists an open set $B$ containing $x$ such that $(B\cap S)-\{x\}=\emptyset$. Moreover, we already know $x\in B\cap S$, we can conclude $B\cap S=\{x\}$. Since $B$ is open, we also have $(B\cap U)-\{x\}\neq\emptyset$, which implies $B\cap U\neq\emptyset$. We already know $S\cap U\neq\emptyset$, hence $B\cap S\cap U\neq\emptyset$, which leads to $\{x\}\cap U\neq\emptyset$, and finally $x\in U$. This means $x\in S\cap U$.

So, if $x$ is a limit point of $U$, $x\in\overline{S\cap U}$.

Now, what if $x\in\overline{U}$ is not a limit point of $U$? In that case, we’re sure $x\in U$, otherwise $x$ is one of the limit points added to the closure. Since $x$ is not a limit point, there exists an open set $B\ni x$ such that $(B\cap U)-\{x\}=\emptyset$. Furthermore, we know that $x\in B$ and $x\in U$, therefore $B\cap U=\{x\}$. Since $S$ is dense and $B,U$ are open sets, $S\cap B$ and $S\cap U$ are non-trivial. This leads to $S\cap B\cap U\neq\empty$ $\Rightarrow$ $S\cap B\cap U = S\cap\{x\}=\{x\}$. Hence $x\in S$, which implies $x\in U\cap S$, and finally $x\in \overline{S\cap U}$.

Hence $\overline{U}\subseteq\overline{S\cap U}$.

Hence $\overline{S\cap U}=\overline{U}$.

## Neighborhood

Definition. Let $(X,\mathcal{T})$ a topo space, N a subset of X, and p a point in N. N is called a neighborhood of p if there exists some open set U such that $p\in U\subseteq N$.

One can have an alternative definition of limit point from this: A point x is a limit point of A iff every neighborhood of x intersect A at a point other than x.

## Connectedness

Given a set $S$ of real numbers. As you may have known, if there exists $b\in S$ such that $b$ is greater than any other numbers of S then b is called the greatest element. S is said to be bounded above if there exists a real number c such that c is greater than any element in S. We call c an upper bound of S. The least upper bound is called supremium. Similarly, the greatest lower bound is called infimum.

Lemma. Let $S\subseteq \mathbb{R}$ and S is bounded above with p being its supremum. If S is closed, then $p\in S$.

Theorem. The only clopen subsets of $\mathbb{R}$ are $\mathbb{R}$ and $\emptyset$.

Definition. Let $(X,\mathcal{T})$ a topo space. It is said to be connected if the only clopen sets are X and $\emptyset$.

Remark. Let $(X,\mathcal{T})$ a topo space. It is said to be disconnected iff there exists a non-empty set A different from X such that A and X-A are open.

The notion of connectedness is very important and we shall discuss in the next posts.

## Reference sources

Special thanks to Tran Hoang Bao Linh for giving some nice examples in this post.