This site is being phased out.

# Open and closed sets

Redirect to:

Suppose the set of all neighborhoods $\gamma$ in $X$ is given. We define open sets as ones where every point has its own neighborhood:

**Definition.** A subset $W$ of $X$ is called *open* (with respect to $\gamma$) if for any $x$ in $W$ there is a neighborhood $U$ ($\in \gamma$) of $x$ that lies entirely within $W$.

For example all "open" intervals, finite or infinite, are open in ${\bf R}$: $$(0,1), (0, \infty ), (- \infty ,0).$$

"Open" disks on the plane, and balls in the Euclidean space are also open.

**Definition.** A set is called *closed* if its complement is open.

For example all "closed" intervals, finite or infinite, are closed in ${\bf R}$: $$[0,1], [0, \infty ), (- \infty ,0].$$

"Closed" disks on the plane, and "closed" balls in the Euclidean space are also closed. Points too.

Some sets are neither closed nor open (unlike doors). Examples in in ${\bf R}$: $$[0,1),$$ $$P = \{ \frac{1}{n} \colon n=1,2, \ldots \}.$$

To see that the latter is neither (with respect to the standard Euclidean topology) consider two facts:

- $x=1$ does not have an interval around it that lies inside $P$;
- $x=0$ is in the complement of $P$ but it does not have an interval around it that lies inside in ${\bf R} \setminus P$.

However $P \cup \{0 \}$ is closed.

**Exercise.** Prove that if $U$ is open with respect to the disks it is also open with respect to the squares, and vice versa. Solution:

**Theorem.** *The intersection of two open sets is open.*

**Proof.** Rewrite the definition three times for each of the three sets involved, compare, then use (B2). More:

**Theorem.** *The union of any collection of open sets is open.*

**Proof.** ** Exercise.** $\blacksquare$

To summarize...

**Theorem.** *The set $\tau$ of all open sets satisfies the following conditions:*

- (T1) $\emptyset, X \in \tau$;
- (T2) if $\alpha \subset \tau$ then $\cup \alpha \in \tau$;
- (T3) if $U,V \in \tau$ then $U \cap V \in \tau$.

**Exercise.** Prove (T1).

**Definition.** A collection $\tau$ that satisfies conditions (T1) - (T3) is called the *topology* of $X$ (with respect to the neighborhoods $\gamma$).

**Corollary.** The topology in ${\bf R}^2$ with respect to the disks is the same as the topology with respect to the squares.

For more details and examples see Neighborhoods and topologies.

**Theorem.** *The union of two closed sets is open.*

**Proof.** Rewrite the definition three times for each of the three sets involved, compare, then use (B2). $\blacksquare$

**Theorem.** *The intersection of any collection of open sets is open.*

**Proof.** **Exercise.** $\blacksquare$

To summarize...

**Theorem.** *The set σ of all closed sets satisfies the following conditions:*

- (T1') $\emptyset, X \in \sigma$;
- (T2') if $\alpha \subset \sigma$ then $\cap \alpha \in \sigma$;
- (T3') if $P,Q \in \sigma$ then $P \cup Q \in \sigma$.

**Proof.** (T3') follows from (T3), illustrated below:

$\blacksquare$

From (T3) and (T3') it follows:

**Corollary.**

- The intersection of a finite number of open sets is open.
- The union of a finite number of closed sets is closed.

**Proof.** **Exercise.** $\blacksquare$

To see why the corollary can't be extended to infinite collections of sets consider these examples:

- $\cap \{( -\frac{1}{n}, \frac{1}{n}): n = 1,2, \ldots \} = \{0 \}$;
- $\cup \{[-1 + \frac{1}{n}, 1 - \frac{1}{n}] \colon n = 1,2, \ldots \} = (-1,1)$;

So, there is no (T3) for closed sets and no (T3') for open sets.

For more examples see Realizations of cubical complexes.