This is my notes for the second chapter of the book “Topology without Tears” by Sidney Morris. We’re going to discuss the Euclidean topology. During the writing of this note, I also had the first sense of the close relationship between geometry and topology. This is one of the rare notes that have GeoGebra visualizations, so keep reading! You will find one in a proof somewhere in this post!

## The Euclidean Topology on \(\mathbb{R}\)

**Definition.** A subset *S* in \(\mathbb{R}\) is said to be open in the Euclidean topology on \(\mathbb{R}\) if for each \(x\in S\), there exist \(a,b\in\mathbb{R}\) such that \(x\in(a,b)\subseteq S\)

It is common that if we say the topology on \(\mathbb{R}\) without defining the topology, we mean the Euclidean topology.

**Properties**

Let \(r,s\in\mathbb{R},\ r < s\).

*(r,s)*, \((r,\infty)\) and \((-\infty, r)\) are open sets.*[r,s]*, \(\{r\}\), \(\mathbb{Z}\), \([r,\infty)\), and \((-\infty, r]\) are closed sets.*[r,s)*,*(r,s]*, \(\mathbb{Q}\), \(\mathbb{I}\) are neither closed nor open.- The only clopen sets are the trivial sets, i.e. \(\mathbb{R},\ \emptyset\).
- Not all open sets are intervals
*(r,s)*. Some open sets may be the union of several intervals.

If *F* is a non-empty countable subset of \(\mathbb{R}\), then *F* is not open, but it can be closed or not closed, depends on the choice of *F*. ▶ Proof

Suppose *F* is open, then for each \(x\in F\), there exist \(a,b\in\mathbb{R}\) such that \(x\in (a,b)\subseteq F\). However, *F* is countable and *(a,b)* is uncountable, contradicts \((a,b)\subseteq F\). Hence *F* is not open.

*F* can be closed or not, dependently on *F* itself. For example, if *F* is the set of all primes, \(\mathbb{R}-F=(-\infty, 2)\cup(2,3)\cup(3,5)\cup\dots\) is the infinite union of open sets, hence an open set.

On the other hand, suppose \(F=\mathbb{Q}\) the set of all rationals. *F* is still countable because it is isomorphic to \(\mathbb{Z}\times\mathbb{Z}\), which is the Cartesian product of two countable sets. However, we cannot use the same argument as presented when *F* is the set of all primes, because we basically cannot “sort” *F* (between any two rational numbers there are infinitely many other rational numbers). We shall check if \(\mathbb{R}-F=\mathbb{I}\) is open using the definition and proof by contradiction. Suppose it is open, then for every \(x\in\mathbb{I}\) we can find an interval \((a,b)\) containing \(x\) and is a subset of \(\mathbb{I}\). However, such an interval will never exist because in the interval there are rational numbers, so it can’t be a subset of \(\mathbb{I}\). Hence \(\mathbb{I}\) is not open and \(F=\mathbb{Q}\) is not closed.

### \(F_{\sigma}\)-set

**Definition.** In a topology space \((X,\mathcal{T})\), a subset *S* is said to be an \(F_{\sigma}\)-set if it is the union of countable number of closed sets.

**Corollary.** In the Euclidean topology on \(\mathbb{R}\), all intervals *(a,b)* and *[a,b]* are \(F_{\sigma}\)-sets.

### \(G_{\delta}\)-set

**Definition.** In a topology space \((X,\mathcal{T})\), a subset *S* is said to be an \(G_{\delta}\)-set if it is the intersection of countable number of open sets.

**Corollary.** In the Euclidean topology on \(\mathbb{R}\), all intervals *(a,b)* and *[a,b]* are \(F_{\sigma}\)-sets.

### Examples

1. Let \(S=\{0, 1, 1/2, 1/3, \dots, 1/n,\dots\}\). Prove that *S* is closed.
▶ Proof

is the union of open intervals, hence open. \(\Rightarrow S\) is closed.

2. Is the set \(S=\{1, 1/2, 1/3, \dots, 1/n,\dots\}\) closed? ▶ Answer

This is not an open set because by definition, when \(x=0\) we cannot find any open interval containing 0 and is a subset of \(\mathbb{R}-S\). Hence \(S\) is not closed. (This is another example of a countable set **not** being a closed set.)

## Basis of a Topology

**Theorem.** A subset *S* in \(\mathbb{R}\) is open iff it is a union of open intervals.

Just like a vector space, in a topological space, the notion “basis” also appears and is defined below:

**Definition.** Let \((X,\mathcal{T})\) be a topo space. A collection \(\mathcal{B}\subseteq\mathcal{T}\) is said to be a basis for the topology \(\mathcal{T}\) if every open set of \(\mathcal{T}\) can be represented as the union of members of \(\mathcal{B}\).

Some examples:

- \(\mathcal{B}=\{(a,b): a,b\in\mathbb{Q}, a<b\}\) is the basis of the Euclidean topo on \(\mathbb{R}\).
- \(\mathcal{B}=\{(a,b): a,b\in\mathbb{R}, a<b\}\) is the basis of the Euclidean topo on \(\mathbb{R}\).
- \(\mathcal{B}=\{\{a\}:a\in X\}\) is the basis of the discrete topo space on
*X*.

By definition, there can be many bases for the same topo. It can easily be seen that if \(\mathcal{B}\subseteq\mathcal{T}\) is a basis, then any \(\mathcal{B}'\) that \(\mathcal{B}\subseteq \mathcal{B}'\subseteq\mathcal{T}\) is also a basis. However, one cannot arbitrarily choose a set \(B\) and generate \(\mathcal{T}\) and call \(\mathcal{T}\) a topology. There are certains conditions so that \(\mathcal{B}\) is a basis.

**Theorem.** Let *X* be a non-empty set and let \(\mathcal{B}\) be a collection of subsets of *X*. The collection of all unions of members of \(\mathcal{B}\) is a topology iff:

- \(X = \bigcup_{B\in\mathcal{B}} B\), and
- \(\forall B_1, B_2\in\mathcal{B},\ B_1\cap B_2\) is a union of members of \(\mathcal{B}\).

If a collection \(\mathcal{B}\) satisfies these conditions, there is a **unique** topology for which \(\mathcal{B}\) is the basis. This topology consists of all unions of members of \(\mathcal{B}\). Otherwise, \(\mathcal{B}\) is not a basis for **any** topology. ▶ Why unique?

Let’s call \(\mathcal{T}_1\) the topology generated by \(\mathcal{B}\) (i.e. \(\mathcal{T}_1\) is the set of all possible unions of members of \(\mathcal{B}\)). Assume that there exists \(\mathcal{T}_2\) a topology in which each open set is a union of members of \(\mathcal{B}\). That means \(\mathcal{B}\subseteq\mathcal{T}_2\subseteq\mathcal{T}_1\). Suppose that \(\mathcal{T}_2\neq\mathcal{T}_1\), so there is a subset of *X* that is a union of members of \(\mathcal{T}_2\) (because \(\mathcal{B}\subseteq\mathcal{T}_2\)), but is not in \(\mathcal{T}_2\), so \(\mathcal{T}_2\) is not a topology. By contradiction, we conclude \(\mathcal{T}_2=\mathcal{T}_1\) and the topology that has \(\mathcal{B}\) as its basis is unique.

If \(\mathcal{B}\) is a basis of \(\mathcal{T}\), then: a subset *S* of *X* is open **iff** *S* is a union of members of \(\mathcal{B}\).

**Sum up:** One topology can have many bases, but a topology is unique to its basis. It is a well-defined surjective mapping from the class of basis to the class of topology.

### Open rectangle

We define an open rectangle (whose sides parallel to the axes) on the plane to be:

\[\{(x,y): (x,y)\in\mathbb{R}^2,\ a<x<b,\ c<y<d\}\subseteq\mathbb{R}^2\]The collection \(\mathcal{B}\) of all open rectangles is the basis for the Euclidean topology on \(\mathbb{R}^2\).

In general, let’s define an n-orthotope (whose sides parallel to the axes) in an \(\mathbb{R}^n\) space to be:

\[\{(x_1,x_2,\dots,x_n):(x_1,x_2,\dots,x_n)\in\mathbb{R}^n,\ a_i<x_i<b_i,\ i=1,2,\dots,n\}\]The collection \(\mathcal{B}\) of all n-orthotopes is the basis for the Euclidean topology on \(\mathbb{R}^n\).

### Open disc

**Theorem.** Every disc \(\{(x,y):(x-a)^2+(y-b)^2<c^2,\ a,b,c\in\mathbb{R}^2\}\) is an open subset of \(\mathbb{R}^2\). ▶ Proof

Consider a disc \(D=\{(x,y):(x,y)\in\mathbb{R}^2,\ x^2+y^2<1\}\). Choose an arbitrary point \((r_x, r_y)\) within this disc. Let \(r=\sqrt{r_x^2+r_y^2}\). Construct an open rectangle with \((r_x\pm\frac{1-r}{2},\ r_y\pm\frac{1-r}{2})\) be the vertices. We need to prove this rectangle lies completely within the disc, i.e. its four vertices lie in the disc.

Let \(d\) be the greatest distance from the center of the disk \((0,0)\) to the rectangle’s vertices. By the triangle inequality, we have \(d\) smaller than the total distance from the center to the point \((r_x, r_y)\) and the distance from there to the furthest vertex.

\[d \lt r+\frac{1-r}{2}\sqrt{2}\]On the other hand, since \(\sqrt{2}/2\lt 1\), we have

\[r+\frac{1-r}{2}\sqrt{2}\lt r+1-r=1\]Hence \(d<1\).

In the visualization below, you can drag the point C around. As long as C lies in the disc, you can see that the rectangle (or more precisely, the square) always lie in the disc.

That means we can say that this disc is composed of the infinite uncountable union of open rectangles, hence it is an open set. In a general case, we can also prove any disc is open by applying similar arguments as above.

**Theorem.** The collection of all open discs is a basis for a topology on \(\mathbb{R}^2\). ▶ Proof idea

It’s easy to see that the union of all open discs form \(\mathbb{R}^2\). We need to prove that the intersection of any two discs can be decomposed into a (uncountable and infinite) union of open discs. We use a similar technique as the proof for the previous theorem. Consider a point \((x,y)\in D_1\cap D_2\). Find a way to draw a disc that takes this point as its center, such that the disc completely lies within \(D_1\cap D_2\). Then we conclude the intersection is composed of an infinite uncountable union of open discs. Hence the set of all open discs is a basis.

**Note:**

- The circumference of a disc is a closed set in \(\mathbb{R}^2\).
- An n-sphere is a closed set in \(\mathbb{R}^{n+1}\).
- An n-ball (the surface of the n-sphere and the space within it) is a closed set in \(\mathbb{R}^{n+1}\)

### Second axiom of countability

If one basis of a topology is countable, we say that the topology is second countable and satisfies the second axiom of countability.

Examples:

- \(\mathbb{R}\) is second countable. The collection of all open intervals \((a, b)\) whereas \(a,b\in\mathbb{Q}\) is the basis of \(\mathbb{R}\). The basis is countable.
- \(\mathbb{R}^n\) is second countable.
- The discrete topology on an uncountable set is
**not**second countable. ▶ Why?

We need to prove that any basis for this topology is uncountable. First, we concede that the singleton sets are open sets, so each must be a union of members of \(\mathcal{B}\) (the basis). That means each singleton set must also appear in \(\mathcal{B}\) itself. The set on which this topology is defined is uncountable, so there are uncountably many singleton sets. \(\Rightarrow \mathcal{B}\) is uncountable.

- The finite-closed topology on \(\mathbb{Z}\) is second countable. ▶ Why?

We first prove that there is countably many finite closed sets. As \(\mathbb{Z}\) is countable, it can be written in the form \(\{z_0, z_1, z_2,\dots\}\). A finite set \(S\subset \mathbb{Z}\) has the form \(\{z_{k_0},z_{k_1},\dots,z_{k_n}\}\) whereas \(k_i\in\mathbb{N},\ i\in\{0,1,\dots,n\}\). We shall have a binary representation for *S*. It is obvious now that there is a one-to-one correspondence between the class of finite subset of \(\mathbb{Z}\) and the class of natural number. Hence the class of finite subset of \(\mathbb{Z}\) is countable. Furthermore, there is also a one-to-one correspondence between the class of closed sets and the class of open sets (by definition *S* is closed iff *X-S* is open). We can conclude that there are countable number of open sets. The basis of the topology is a subset of the topology itself, so the basis is also countable. Hence the topology satisfies the second axiom of countability.

## Basis for a given topology

**Theorem.** Let \((X,\mathcal{T})\) be a topological space. A set \(\mathcal{B}\subseteq\mathcal{T}\) is a basis of \(\mathcal{T}\) **iff** \(\forall x\in U\in \mathcal{T}\), there exist \(B\in\mathcal{B}\) such that \(x\in B\subseteq U\). ▶ Proof

1. Prove that if \(\mathcal{B}\) is a basis of \(\mathcal{T}\), then \(\forall x\in U\in \mathcal{T}\), there exist \(B\in\mathcal{B}\) such that \(x\in B\subseteq U\).

Since *U* is an open set, it is the union of \(B_i,\ i\in I\), where \(B_i\in\mathcal{B}\) and \(I\) is some index set. But \(x\in U\) implies that there exists some \(B_i\) such that \(x\in B_i\), and obviously \(B_i\subseteq U\).

2. Prove that \(\mathcal{B}\) is a basis of \(\mathcal{T}\) if \(\forall x\in U\in \mathcal{T}\) there exist \(B\in\mathcal{B}\) such that \(x\in B\subseteq U\).

This case is pretty obvious, any open set can be written:

\[U=\bigcup_{x\in U}B_x\]where \(B_x\) is the member of \(\mathcal{B}\) that contains the point *x*.

**Theorem.** Let \(\mathcal{B}\) be a basis of the topo space \((X,\mathcal{T})\). A subset *U* of *X* is an open set iff \(\forall x\in U\), there exists \(B\in\mathcal{B}\) such that \(x\in B\subseteq U\). ▶ Proof

1. Prove that if \(U\) is open then \(\forall x\in U\), there exists \(B\in\mathcal{B}\) such that \(x\in B\subseteq U\).

This is basically the same as the first part of the previous theorem’s proof.

2. Prove that \(U\) is open if \(\forall x\in U\), there exists \(B\in\mathcal{B}\) such that \(x\in B\subseteq U\).

Again, very similar to the previous proof, we can show that \(U\) is the union of members of \(\mathcal{B}\), and therefore open.

To sum up: Given a topo space \((X,\mathcal{T}),\ \mathcal{B}\subseteq\mathcal{T},\ U\subseteq X\)

- \(\mathscr{A}\) is the proposition “\(\mathcal{B}\) is a basis”
- \(\mathscr{B}\) is the proposition “
*U*is an open set” - \(\mathscr{C}\) is the proposition “\(\forall x\in U\), there exists \(B\in\mathcal{B}\) such that \(x\in B\subseteq U\)”

We have:

- \(\mathscr{A}\Rightarrow\mathscr{B}\textrm{ and }\mathscr{C}\)
- \(\mathscr{A}\textrm{ and }\mathscr{C}\Rightarrow\mathscr{B}\)
- \(\mathscr{A}\textrm{ and }\mathscr{B}\Rightarrow\mathscr{C}\) (note: must consider all open sets)
- \(\mathscr{B}\textrm{ and }\mathscr{C}\Rightarrow\mathscr{A}\)

### When do two bases generate the same topology?

**Theorem.** Let \(\mathcal{B}_1, \mathcal{B}_2\) be the bases for topologies \(\mathcal{T}_1, \mathcal{T}_2\), respectively, on a non-empty set *X*. Then \(\mathcal{T}_1=\mathcal{T}_2\) iff:

- \(\forall x,B:x\in B\in\mathcal{B}_1\), there exists \(B'\in\mathcal{B}_2\) such that \(x\in B'\subseteq B\). (i.e. \(\mathcal{B}_2\) is a basis of \(\mathcal{T}_1\))
- \(\forall x,B:x\in B\in\mathcal{B}_2\), there exists \(B'\in\mathcal{B}_1\) such that \(x\in B'\subseteq B\). (i.e. \(\mathcal{B}_1\) is a basis of \(\mathcal{T}_2\))

## Subbasis

**Definition.** Let \((X,\mathcal{T})\) be a topo space. A non-empty collection \(\mathcal{S}\) of subsets of *X* is called a subbasis if the collection of all finite intersections of memebers of \(\mathcal{S}\) forms a basis of \(\mathcal{T}\).

Examples:

- The collection of all open intervals \((-\infty, a)\) and \((b,\infty)\) is a subbasis for (the Euclidean topo) \(\mathbb{R}\).
- The collection of all closed intervals \([a,b]\) where \(a\lt b\) is a subbasis for the discrete topology on \(\mathbb{R}\).
- The collection of all sets \(X-\{x\},\ x\in X\) is a subbasis for the finite-closed topology on
*X*, where*X*has at least 2 points. ▶ Proof idea

We see that the finite intersection of some sets \(X-\{x_i\},\ i\in\{1,\dots,n\}\), \(x_i\in X\) is \(X-\{x_1,x_2,\dots,x_n\}\), which is an open set of the finite-closed topology on *X*. Let \(\mathcal{B}\) be the collection of all such intersections. We concede that \(\mathcal{B}\) contains all open sets of *X*, except that \(X\notin\mathcal{B}\). However, \(\mathcal{B}\) covers *X*, hence any open set is a union of members of \(\mathcal{B}\). And by definition of a basis, we conclude \(\mathcal{B}\) is a basis of the finite-closed topo.