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 \(a<b,\ a,b\in\mathbb{R}\) 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*:

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.