In the lesson Introduction to set theory you learned about set notation and various properties of sets. In addition, in this lesson we will talk about some specific sets that are subsets of a certain metric space.

## Open ball

**Definition:** Let $x \in \mathbf{R^{n}}$ and $r>0$. A set

**$$K(x, r) = \{ y \in \mathbf{R^{n}}: d(x,y)<r\} = \left \{y \in \mathbf{R^{n}} : \sqrt{\sum_{i=1}^{n}(x_{i} – y_{i})^{2}}< r \right \}$$**

is called an **open ball** with center *x* and radius *r*.

**N****o****te****:**Expression $d(x, y)$ represents the **distance function**. In other words, *d* is the **standard Euclidean distance** between two points.

Pictures below represent open balls in sets $\mathbf{R}, \mathbf{R^{2}}$ and $\mathbf{R^{3}}$, respectively.

## Open sets

**Definition:** We say that a set $A \subseteq \mathbf{R^{n}}$ is an **open** **set** if

**$$(\forall x \in A) \ (\exists r>0) \ K(x, r) \subseteq A.$$**

In other words, **every open set is the union of open balls**.

You can think of it as a collection of elements that doesn’t include any limit points.

**Example 1:** An open ball $K(x, r)$ is an open set.

**Example 2:** An open interval $\left<0, 1\right>$ is an open set in $\mathbf{R}$, but not in $\mathbf{R^{2}}$. More precisely, it is not an open set in $\mathbf{R^{2}}$ because we identify it with a set $\{(x, 0): 0< x < 1 \subset \mathbf{R^{2}}\}$.

Moreover, any open interval is an open set.

**Example 3:** $\emptyset$ and $\mathbf{R^{n}}$ are open sets.

## Topology on a set

**Theorem:** Let *X* be a metric space. Furthermore, let $\mathcal{T}$ be a family of all open sets with properties:

(1) $\emptyset, X \in \mathcal{T}$

(2) the union of any family from $\mathcal{T}$ is from $\mathcal{T}$, i.e.

$$U_{\alpha}\in \mathcal{T} \rightarrow \bigcup U_{\alpha}\in \mathcal{T}$$

(3) the intersection of any finite family of elements from $\mathcal{T}$ is an element from $\mathcal{T}$, i.e.

$$U_{i} \in \mathcal{T}, \ i = 1, \cdots, m, \rightarrow \bigcap_{i=1}^{m} U_{i} \in \mathcal{T}.$$

**Definition:** We say that $\mathcal{T}$ is a** topological structure** or simply a **topology** on set X. Furthermore, ordered pair **$(X, \mathcal{T})$** is called a **topological space**.

Elements of a set *X* are **points** and elements of a family $\mathcal{T}$ are **open sets** of the topological space $(X, \mathcal{T})$.

**Example 4:** Let $X = \{1, 2, 3\}$ and $\mathcal{T} = \{\emptyset, X, \{1, 2\}, \{1, 3\}\}$. Is $\mathcal{T}$ a topology on *X*?

**Solution: **$\mathcal{T}$ is not a topology on *X *because $\{1, 2\} \cap \{1, 3\} = \{1\} \notin X$.

## Interior of a set

**Definition:** Let $A \subseteq \mathbf{R}$. A point $x \in A$ is the **interior point** of a set *A* if there exists an open set *U* such that $x \in U \subseteq A$.

The **interior** of a set *A* is a set of all interior points. In other words,

$$IntA = \{x \in \mathbf{R^{n}}: \exists r >0, \ K(x, r) \subseteq A\}.$$

**Example 5:** *Int A* is an open set. Moreover, it is the largest open set in *A*.

**Example 6:** Interior of a set $S = \{(x, y) \in \mathbf{R}^{2} : 0< x \leq 1\}$ is $Int S = \{(x, y) \in \mathbf{R}^{2} : 0 < x < 1\}$.

Namely, we can put all points $(x, y)$ (for which $0 < x < 1$ is valid) in an open set in *S*, so all that points are in *Int S*. On the other side, for a point $(1, y)$ every open set of radius *r* contains a point which is not an element of *S*. For instance, point $\left (1 +\frac{r}{2}, y \right)$. Therefore, every open set which contains a point $(1, y)$, also contains a point which is not in *S*.