# Math 4201 Topology I (Lecture 7)

## Review from last lecture

Not every open set in subspace topology is open in the original space

Let $X=\mathbb{R}$ with standard topology and $Y=[0,1]\cup [2,3]$. equipped with subspace topology generated by the standard basis for $\mathbb{R}$.

so $[0,1]=(-1,\frac{3}{2})\cap Y$ In particular, $[0,1]$ is open set in $Y$, but not an open set in $\mathbb{R}$.

## New materials

### Closed sets in topological space

#### Proposition of open set in subspace topology

If $X$ is a topological space, then $Y\subseteq X$ is open and is with the subspace topology. If $Z\subset Y$ is open subspace of $Y$, then $Z$ is also an open subspace of $X$.

<details>
<summary>Proof</summary>

Since $Z\subset Y$ is open in the subspace topology, there is an open $U\subset Y$ such that $Z=U\cap Y$.

SInce $Z$ is the intersection of open sets in $X$, then $Z$ is open in $X$.
</details>


#### Definition of closed set

For any topology $\mathcal{T}$ on a set $X$, a subset $Z\subseteq X$ is said to be **closed** if its complement $Z^c$ is an open set (in $X$).

> Note the complement is defined $Z=X\setminus Z$.

<details>
<summary>Example of closed set in standard topology of real numbers</summary>

For example, $[a,b]$ is a closed set in the standard topology of real numbers. since $\mathbb{R}-[a,b]=(-\infty,a)\cup (b,\infty)$ is an open set.
</details>

<details>
<summary>Example of closed set in trivial topology</summary>

For any set $X$, the trivial topology is $\mathcal{T}_0=\{\emptyset, X\}$.
Since $X^c=\emptyset$ is an open set, $X$ is a closed set.
Since $\emptyset^c=X$ is an open set, $\emptyset$ is a closed set.

</details>

<details>
<summary>Example of closed set in finite complement topology</summary>

For any set $X$, the finite complement topology is $\mathcal{T}_1=\{U\subseteq X\mid X\setminus U\text{ is finite}\}$.

Then the set of all finite subsets of $X$ is a closed set.
</details>

For general, if $\mathcal{T}$ is a topology on $X$, then:

1. $\emptyset, X$ are closed sets.
2. $\mathcal{T}$ is closed with respect to arbitrary unions.

Let $\{Z_\alpha\}_{\alpha \in I}$ be an arbitrary collection of closed sets in $X$. Then $X-Z_\alpha$ is open for each $\alpha \in I$. So $\forall \alpha \in I$. $\bigcup_{\alpha \in I} (X-Z_\alpha)=X-\bigcap_{\alpha \in I} Z_\alpha$ is open. So $\bigcap_{\alpha \in I} Z_\alpha$ is closed.

So the corollary is: an arbitrary intersection of closed sets is closed.

3. $\mathcal{T}$ is closed with respect to finite intersections. This also implies that a finite union of closed sets is closed.

If $\{Z_1, Z_2, \ldots, Z_n\}$ is a closed subset of $X$, then $X-Z_i$ is open for each $i=1,2,\ldots,n$. So $\forall i=1,2,\ldots,n$. $\bigcap_{i=1}^n (X-Z_i)=X-\bigcup_{i=1}^n Z_i$ is open. So $\bigcup_{i=1}^n Z_i$ is closed.

> [!NOTE]
>
> We can also define the topology using the closed sets instead of the open sets.
>
> 1. $\emptyset, X$ are closed sets.
> 2. $\mathcal{T}$ is closed with respect to arbitrary intersections.
> 3. $\mathcal{T}$ is closed with respect to finite unions.
>
> This yields the same topology.

#### Theorem of closed set in subspace topology

Let $X$ is a topological space and $Y\subseteq X$ equipped with the subspace topology.

A subset $Z\subseteq Y$ is closed in $Y$ if and only if $Z$ is the intersection of a closed $W\subseteq X$ and $Y$. That is $Z=W\cap Y$.

<details>
<summary>Proof</summary>

$\Rightarrow$

If $Z$ is closed in $Y$, then $Y-Z$ is open in $Y$.

Then, there is open set $U\subseteq X$ such that $Y-Z=U\cap Y$.

So $Z=(X-U)\cap Y$, $X-U$ is closed in $X$ because $U$ is open in $X$.

Take $W=X-U$.

$\Leftarrow$

If $Z=W\cap Y$ for some closed $W\subseteq X$, then $Y-Z=Y-(W\cap Y)=(Y-W)\cap Y$ is open in $Y$.
So $Z$ is closed in $Y$.

</details>

#### Lemma of closed in closed subspace

Let $X$ is a topological space and $Y\subseteq X$ is closed and is equipped with the subspace topology. Then any closed subset of $Y$ is also closed in $X$.

> [!WARNING]
>
> Not any subset of a topological space $X$ is either open or closed.

<details>
<summary>Example of open and closed subset</summary>

Let $X=\mathbb{R}$ with standard topology.

$(a,b)$ is open, but not closed.
$[a,b]$ is closed, but not open.
$[a,b)$ is neither open nor closed.
$\emptyset,\mathbb{R}$ is both open and closed.

</details>

<details>
<summary>Example of open and closed subset in other topologies</summary>

Let $X=[0,1]\cup (2,3)$ induced by the standard topology of $\mathbb{R}$.

$Z=[0,1]$ is an open subset of $X$.

$Z=[0,1]$ is also closed subset of $X$ since $Z=[0,1]\cap X$ is open in $\mathbb{R}$.

</details>

We can associate an open and a closed to any subset $A$ of a topological space $X$.

#### Interior and closure of a set

The interior of a set $A$ is defined as follows:

$$
\operatorname{Int}(A)=\bigcup_{U\subseteq A, U\text{ is open in }X} U
$$

Also denoted as $A^\circ$.

The interior of a set $A$ is the largest open subset of $A$.

That is $\forall U\subseteq A, U\text{ is open in }X$, then $U\subseteq \operatorname{Int}(A)$. (by definition that $U$ must be in collection of open sets that is a subset of $A$)

#### Closure of a set

The closure of a set $A$ is the smallest closed subset of $X$ that contains $A$.

> Note that if we change the definition as the intersection of all closed subsets of $X$ that **contained in $A$**, we will get the empty set.

$$
\overline{A}=\bigcap_{A\subseteq C, C\text{ is closed in }X} C
$$

The closure of a set $A$ is the smallest closed subset of $X$ that contains $A$. (follows the same logic as the previous definition)