Equivalence Between an Open Set and Its Interior
A set \( A \) in a topological space \( X \) is open if and only if it coincides with its interior. $$ A = \text{Int}(A) $$
In other words, \( A \) is open if every point in \( A \) has an open neighborhood that is entirely contained within \( A \).
Therefore, \( A \) is open if and only if \( A = \text{Int}(A) \), meaning it contains all the open sets that it possibly can.
The interior of a set Int(A) is the largest open set contained within \( A \), which is the union of all open sets within \( A \).
A Practical Example
Consider the topological space \( \mathbb{R} \) with the standard topology, where every open interval is an open set.
Let's examine some sets and verify if they are open using the characterization \( A = \text{Int}(A) \).
Example 1
Consider the open interval \( A = (0, 1) \)
$$ A = (0, 1) $$
The interior of \( A \) is the set (0,1).
$$ \text{Int}(A) = (0,1) $$
Since the set \( A \) coincides with its interior, we can deduce that \( A \) is an open set.
Example 2
Consider the closed interval \( B = [0,1] \)
$$ B = [0, 1] $$
The interior of the set is the interval from 0 to 1, excluding the endpoints.
$$ \text{Int}(B) = (0,1) $$
In this case, the set \( B = [0,1] \) does not coincide with its interior \( \text{Int}(B) = (0,1) \), so we deduce that \( B \) is not an open set.
Note: These practical examples demonstrate how the definition of a set's interior can be used to verify if the set is open or not.
The Proof
We need to prove the equivalence between an open set \( A \) and its interior \( \text{Int}(A) \).
We divide the proof into two parts:
1] If \( A \) is open, then \( \text{Int}(A) = A \)
Assume that \( A \) is an open set.
By definition, the interior \(\text{Int}(A)\) is the set of all points in \( A \) that have an open neighborhood contained within \( A \).
Since \( A \) is open, every point \( x \in A \) has an open neighborhood \( U \subseteq A \).
Therefore, every point in \( A \) is also a point in \(\text{Int}(A)\).
$$ A \subseteq \text{Int}(A) $$
On the other hand, by definition, the interior of \( A \) is the union of open sets contained within \( A \).
$$ \text{Int}(A) \subseteq A $$
Thus, this double inclusion implies that the two sets are equal:
$$ A = \text{Int}(A) $$
2] If \( A = \text{Int}(A) \), then \( A \) is open
Suppose that \( A = \text{Int}(A) \).
We need to show that \( A \) is open.
Take any point \( x \in A \).
Since \( x \in \text{Int}(A) \) and \( \text{Int}(A) = A \), by the definition of the interior, \( x \) must have an open neighborhood \( U \subseteq \text{Int}(A) = A \).
This implies that every point in \( A \) has an open neighborhood contained within \( A \).
Therefore, \( A \) is an open set.
3] Conclusion
We have thus shown that a set \( A \) in a topological space \( X \) is open if and only if \( A \) coincides with its interior, i.e., \( A = \text{Int}(A) \).
And so on.
