Inclusion Property of Closure in a Closed Set

If $C$ is a closed set in the topological space $X$ and the set $A$ is contained within $C$, then the closure of $A$, denoted as $\text{Cl}(A)$, is a subset of $C$.   $$A \subseteq C \ , \ C \text{ is closed } \implies \operatorname{Cl}(A) \subseteq C$$

This is because the closure of $A$ is the smallest closed set that contains $A$, and since $C$ is already closed and contains $A$, it must also contain $\text{Cl}(A)$.

Therefore, $\text{Cl}(A)$ cannot extend beyond $C$.

A Practical Example

Consider the topological space $X = \mathbb{R}$ of real numbers with the standard topology.

In the standard topology, open sets are open intervals.

Take, for example, the set $C = [0, 2]$, which is a closed set in $\mathbb{R}$.

$$C = [0,2]$$

Now, choose a subset, such as the open set $A = (0, 1)$, which is a subset of $C$.

$$A = (0,1)$$

Next, let's determine the closure of the set $A$:

The closure of $A$, $\operatorname{Cl}(A)$, is the smallest closed set in $\mathbb{R}$ that contains $A$.

Therefore, the closure of $(0, 1)$ is $[0, 1]$, since $[0, 1]$ is the smallest closed set that includes all points of $(0, 1)$ along with its accumulation points (0 and 1).

$$\text{Cl}(A) = [0,1]$$

Since $A$ is a subset of the closed set $C$

$$A = (0, 1) \subseteq C = [0, 2]$$

According to the property, the closure $\operatorname{Cl}(A)$ must be contained within $C$.

$$\text{Cl}(A) \subseteq C$$

In fact, $\operatorname{Cl}(A) = [0, 1]$, and we can immediately verify that $[0, 1] \subseteq [0, 2]$.

$$\text{Cl}(A) = [0,1] \subseteq [0,2] = C$$

The result is $(0, 1) \subseteq [0, 2]$, and the closure $\operatorname{Cl}((0, 1)) = [0, 1]$ is indeed a subset of $[0, 2]$.

This practical example demonstrates that if $A$ is a subset of a closed set $C$ in a topological space $X$, then the closure of $A$ is contained within $C$.

The Proof

By definition, $C$ is closed in $X$, so its complement $X \setminus C$ is an open set in $X$.

We already know from the hypothesis that $A \subseteq C$.

The closure of $A$, denoted as $\operatorname{Cl}(A)$, is the smallest closed set in $X$ that contains $A$.

In other words, $\operatorname{Cl}(A)$ is the intersection of all closed sets in $X$ that contain $A$.

Since $C$ is a closed set in $X$ and contains $A$, $C$ is one of the sets in the collection of closed sets that contain $A$.

The closure $\operatorname{Cl}(A)$, being the intersection of all closed sets that contain $A$, will also be contained within $C$, since $C$ is one of those sets.

Therefore, we can conclude that the closure of $A$ is a subset of the closed set $C$

$$\operatorname{Cl}(A) \subseteq C$$

Simply put, since $C$ is closed and contains $A$, and because the closure of $A$ is the smallest closed set containing $A$, the closure of $A$ must necessarily be contained within $C$.

And so on.