Inclusion of Open Sets in the Interior of a Set

If $ U $ is an open set in a topological space $ X $ and $ U $ is contained within $ A $, then $ U $ is included in the interior of $ A $. $$ U \subseteq \text{Int}(A) $$

The interior of $ A $, $\text{Int}(A)$, is the largest open set contained within $ A $.

Therefore, any open set $ U $ that is inside $ A $ will also be part of the interior $\text{Int}(A)$ of the set $ A $.

$$ \text{Int}(A) = \bigcup \{ O \subseteq A \mid O \text{ is open in } X \} $$

Among these open sets is the set $ U $ because, by our initial assumption, it is an open set contained within $ A $.

A Practical Example
The Proof

A Practical Example

Consider two sets, $ U $ and $ A $, in the topological space $ \mathbb{R} $ (the real numbers) with the standard topology, where open sets are open intervals and their arbitrary unions.

$$ U = (1, 2) $$

$$ A = [0, 3] $$

The set $ U = (1, 2) $ is open because it is an open interval in $\mathbb{R}$, making it an open set in the standard topology of $\mathbb{R}$.

Furthermore, the set $ U $ is contained within $ A $ because $ U = (1, 2) \subseteq A = [0, 3] $. In other words, every point in $ U $ is also a point in $ A $, so $ U \subseteq A $.

The interior of $ A = [0, 3] $, denoted as $\text{Int}(A)$, is the largest open set contained within $ A $.

In this case, the interior of $ A $ is $(0, 3)$, because $(0, 3)$ is the largest open interval contained within $[0, 3]$.

$$ \text{Int}(A) = (0,3) $$

Since $ U = (1, 2) $ and $\text{Int}(A) = (0, 3)$, it is clear that $ U $ is a subset of the interior of $ A $.

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

In this example, I have demonstrated that $ U $ is an open set in $\mathbb{R}$ and $ U \subseteq A $, hence $ U \subseteq \text{Int}(A) $.

This confirms that if $ U $ is an open set in $ \mathbb{R} $ and $ U \subseteq A $, then $ U \subseteq \text{Int}(A) $.

The Proof

Let $ X $ be a topological space, $ U $ an open set in $ X $, and $ A \subseteq X $ such that $ U \subseteq A $.

By hypothesis:

$ U $ is an open set in $ X $.
$ U $ is contained within $ A $, $ U \subseteq A $.

According to the definition of interior, $ \text{Int}(A) $ is defined as the largest open set contained within $ A $.

Since $ U $ is open in $ X $ and $ U \subseteq A $, $ U $ is one of the open sets that form the interior $ \text{Int}(A) $.

By definition, the interior $ \text{Int}(A) $ is the union of all these open sets contained within $ A $. Since $ U $ is one of these sets, it follows that $ U \subseteq \text{Int}(A) $.

Therefore, if $ U $ is an open set in $ X $ and $ U \subseteq A $, then $ U \subseteq \text{Int}(A) $.

And so forth.