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,

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 $% $ 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:

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 $A\cap B\subseteq\overline{A}\cap\overline{B}$.

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.