The Monotonic Property of Set Closure

The monotonic property of set closure states that if \( A \) and \( B \) are any two sets (not necessarily closed) and \( A \) is a subset of \( B \), then the closure of \( A \) is a subset of the closure of \( B \): \[ A \subseteq B \implies \text{Cl}(A) \subseteq \text{Cl}(B) \]

This concept is so straightforward that it almost goes without saying.

Think of it like this: if you take a small box and close it, then place it inside a larger box and close the larger box, the small box remains closed inside the larger box.

A Concrete Example

Let's use a familiar topological space: the real line \(\mathbb{R}\) with the standard topology.

In real topology, open sets are the open intervals.

Consider the following sets in \(\mathbb{R}\):

\[ A = (0, 1) \]

\[ B = [0, 2] \]

It's clear that \( A \) is a subset of \( B \) because every element of \( A \) is also an element of \( B \).

\[ A \subseteq B \]

The Closure of Set \(A\)

The set \( A \) is the open interval \( (0, 1) \).

The closure of \( A \) is \( A \) combined with its limit points.

The only limit points of \( A \) are \( 0 \) and \( 1 \), because every neighborhood of \( 0 \) and \( 1 \) contains points of \( A \).

Therefore, the closure of \( A \) is:

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

The Closure of Set \(B\)

The set \( B \) is the closed interval \([0, 2]\).

The closure of \( B \) is \( B \) itself, since \( B \) is already closed and includes all its limit points.

\[ \text{Cl}(B) = [0, 2] \]

In Conclusion

Knowing that \( \text{Cl}(A) = [0, 1] \) and \( \text{Cl}(B) = [0, 2] \), it is evident that the closure of \( A \) is a subset of the closure of \( B \):

\[ \text{Cl}(A) \subseteq \text{Cl}(B) \]

In fact, the interval \([0, 1]\) is a subset of the interval \([0, 2]\), confirming that the closure of \( A \) is a subset of the closure of \( B \).

This confirms the monotonic property of set closure.

The Proof

Assume initially that set \( A \) is a subset of \( B \):

\[ A \subseteq B \]

I need to prove that \( \text{Cl}(A) \subseteq \text{Cl}(B) \).

Since \( A \subseteq B \), it is obvious that all points of \( A \) are also points of \( B \).

A point \( x \in \text{Cl}(A) \) if and only if every neighborhood of \( x \) contains at least one point of \( A \) (different from \( x \) itself if \( x \in A \)).

Since \( A \subseteq B \), every neighborhood of \( x \) that contains a point of \( A \) will also contain a point of \( B \), because \( A \) is contained within \( B \).

The closure of a set \( X \) is the intersection of all closed sets containing \( X \).

Since \( A \subseteq B \), any closed set containing \( B \) will also contain \( A \).

Therefore, the intersection of all closed sets containing \( B \) will include all limit points of \( A \) as well.

Thus, \( \text{Cl}(A) \subseteq \text{Cl}(B) \) because every point of \( A \) is also a point of \( B \), and every limit point of \( A \) is also a limit point of \( B \) due to the inclusion of \( A \) in \( B \).

In conclusion, if \( A \subseteq B \), then the closure of \( A \) is a subset of the closure of \( B \), that is \( \text{Cl}(A) \subseteq \text{Cl}(B) \).

This property directly follows from the definition of closure and the inclusion relationship between \( A \) and \( B \).

And so on.