Inclusion of Set Interiors in Topology

If a set $ A $ is contained within a set $ B $, then the interior of $ A $ is necessarily contained within the interior of $ B $. $$ A \subseteq B \Rightarrow \text{Int}(A) \subseteq \text{Int}(B) $$

This result follows from the fact that every open set contained in $ A $ is also contained in $ B $.

Therefore, the operation of taking the interior preserves the inclusion between sets.

A Practical Example
The Proof

A Practical Example

Consider two sets $ A $ and $ B $ in $ \mathbb{R} $ with the standard topology.

$$ A = [1, 3] $$

$$ B = [0, 4] $$

Clearly, set $ A $ is a subset of set $ B $.

$$ A \subseteq B $$

The interior of a set in $ \mathbb{R} $ with the standard topology is the union of all its open subsets.

Interior of A
The set $ A = [1, 3] $ contains the open interval $ (1, 3) $. Thus: \[
\text{Int}(A) = (1, 3)
\]
Interior of B
The set $ B = [0, 4] $ contains the open interval $ (0, 4) $. Thus: \[
\text{Int}(B) = (0, 4)
\]

It is evident that the interior $ \text{Int}(A) = (1, 3) $ is also a subset of the interior $ \text{Int}(B) = (0, 4) $.

$$ \text{Int}(A) \subseteq \text{Int}(B) $$

This example clearly demonstrates that $ A \subseteq B $ implies $ \text{Int}(A) \subseteq \text{Int}(B) $ in $ \mathbb{R} $ with the standard topology.

The Proof

Consider two sets $ A $ and $ B $ in a topological space $ X $ such that $ A \subseteq B $.

We need to show that $ \text{Int}(A) \subseteq \text{Int}(B) $, where $ \text{Int}(A) $ denotes the interior of $ A $.

The interior of a set $ A $, denoted by $ \text{Int}(A) $, is defined as the union of all open sets contained in $ A $.

In other words, $ \text{Int}(A) $ is the largest open set contained within $ A $.

By definition, every open set contained in $ A $ is also a subset of $ B $, since $ A \subseteq B $.

Thus, every open set contained in $ A $ is also contained in $ B $.

Since $ \text{Int}(A) $ is the union of all open sets contained in $ A $ and all these open sets are also contained in $ B $, it follows that $ \text{Int}(A) $ is an open set contained in $ B $.

Furthermore, the interior of $ B $, $ \text{Int}(B) $, is by definition the largest open set contained within $ B $.

Since $ \text{Int}(A) $ is an open set contained within $ B $, it follows that $ \text{Int}(A) $ must be contained within $ \text{Int}(B) $.

In conclusion, if $ A \subseteq B $, then $ \text{Int}(A) \subseteq \text{Int}(B) $.

This demonstrates that taking the interior of sets preserves their inclusion relationship.

And so on.