Note: When \( X \) is known by context then we may say \( A \) is closed to mean that \( A \) is closed in \( X \)
Note: When \( X \) is known by context then we may say \( A \) is closed to mean that \( A \) is closed in \( X \)
The complement of \( \emptyset , X \) in \( X \) are \( X , \emptyset \) respectively, which are both open in \( X \), making the original sets closed
Suppose that \( \left \lbrace A_{\alpha} \right \rbrace_{\alpha \in I} \) is a collection of closed sets, then we know that \( X \backslash \bigcap_{\alpha \in I} A_{\alpha} = \bigcup_{\alpha \in I} \left ( X \backslash A_{\alpha} \right ) \), and since each \( A_{\beta} \) is closed then \( X \backslash A_{\beta} \) is open, making \( \bigcup_{\alpha \in I} \left ( X \backslash A_{\alpha} \right ) \) an arbitrary union of open sets, and is thus open, making \( X \backslash \bigcap_{\alpha \in I} A_{\alpha} \) open
Now suppose that \( I \) is finite, and we'll show that \( X \backslash \bigcup_{\alpha \in I} A_{\alpha} \) is open, we know this set is equal to \( \bigcap_{\alpha \in I} \left ( X \backslash A_{\alpha} \right ) \), and as a finite intersection of open sets, we know it is open as well showing the original set is open
Let's start by assuming that we have a set \( A = C \cap Y \) where \( C \) is closed in \( X \), therefore \( X \backslash C \) is open in \( X \), therefore \( Y \cap \left ( X \backslash C \right ) \) is open in \( Y \)
Note that \( Y \cap \left ( X \backslash C \right ) \) \( = \) \( \left ( X \cap Y \right ) \backslash C \) \( = \) \( \left ( X \cap Y \right ) \backslash \left ( C \cap Y \right ) \) \( = \) \( Y \backslash A \) (where the last step is justified since we know \( Y \subseteq X \) ), this shows that \( Y \setminus A \) is open, meaning that \( A \) is closed in \( Y \)
Now let's assume that \( A \) is closed in \( Y \), therefore \( A \subseteq Y \) and \( Y \backslash A \) is open in \( Y \), meaning that there is some set \( U \) open in \( X \) such that \( Y \backslash A = Y \cap U \)
Let's note that \( Y \setminus \left ( Y \setminus A \right ) \) \( = \) \( \left ( Y \cap A \right ) \cup \left ( Y \setminus Y \right ) = A \cup \emptyset = A\) so that
\[ \begin{align*} A &= Y \setminus \left ( Y \setminus A \right ) \\ &= Y \setminus \left ( Y \cap U \right ) \\ &= \left ( Y \setminus Y \right ) \cup \left ( Y \setminus U\right ) \\ &= Y \setminus U \\ &= \left ( X \cap Y \right ) \setminus U \\ &= Y \cap \left ( X \setminus U \right) \end{align*} \]Since \( X \setminus U \) is closed in \( X \) this shows that \( A \) equals the intersection of a closed set with \( Y \) as needed.
Suppose that \( \mathcal { A } \) is the collection of open sets which are also subsets of \( A \), since \( A \) is open then \( A \in \mathcal { A } \), then \( \overline{ A } = \bigcup \mathcal{ A } = \bigcup \left( \mathcal{ A } \setminus \left\{ A \right\} \right) \cup A \), since each element in \( \mathcal{ A } \setminus \left\{ A \right\} \) is a subset of \( A \), then so is the union, thus \( \bigcup \left( \mathcal{ A } \setminus \left\{ A \right\} \right) \cup A = A \) as needed.
The other direction is easier, suppose that \( A = \operatorname{ Int } \left( A \right) \), then \( \operatorname{ Int } \left( A \right) \) is open, then so is \( A \) as needed.
Suppose that \( \mathcal{ A } \) is the collection of closed sets which are also supersets of \( A \), then \( \overline{ A } = \bigcap \mathcal{ A } = \bigcap \left( \mathcal{ A } \setminus \left\{ A \right\} \right) \cap A \), and since each element in \( \mathcal{ A } \setminus \left\{ A \right\} \) is a superset of \( A \), then so is it's intersection, therefore \( \bigcap \left( \mathcal{ A } \setminus \left\{ A \right\} \right) \cap A = A \)
Now suppose that \( A = \overline{ A } \), since \( \overline{ A } \) is closed then we know \( A \) is closed
Set \( B \) equal to the closure of \( A \) in \( Y \), since \( \bar{A} \) is closed in \( X \), then we know that \( \bar{A} \cap Y \) is closed in \( Y \), also note that \( A \subseteq \bar{A} \cap Y \) because we know \( A \subseteq \bar{A} \) and \( A \subseteq Y \) so we can conclude \( B \subseteq \left( \bar{A} \cap Y \right) \)
On the other hand, we know that \( B \) is closed in \( Y \), thus \( B = C \cap Y \) for some set \( C \) closed in \( X \), we recall that \( A \subseteq B \) so that \( A \subseteq \left( C \cap Y \right) \), therefore \( A \subseteq C \) showing that \( C \) is a closed set containing \( A \), thus \( \bar{A} \subseteq C \) so that \( \left( \bar{A} \cap Y \right) \subseteq \left( C \cap Y \right) = B \), as needed.
We prove the contrapositive, that is \( x \notin \bar{A} \) iff there is some open set \( U \) containing \( x \) that doesn't intersect \( A \)
Suppose that \( x \notin \bar{A} \), since \( \bar{A} \) is closed, then \( U = X \setminus \bar{A} \) must be open and containing \( x \) and since \( A \subseteq \bar{A} \), we know \( U \) doesn't intersect \( A \), as needed.
Now the reverse direction: assuming that we have some open set \( U \) containing \( x \) that doesn't intersect \( A \), we have \( A \subseteq X \setminus U \) and \( X \setminus U \) is closed therefore \( \bar{A} \subseteq X \setminus U \), so that \( \bar{A} \cap U = \emptyset \), since we know \( x \in U \) we know that \( x \notin \bar{A} \) as needed.
Suppose that \( x \in \bar{A} \), therefore every open set containing \( x \) intersects \( A \), now suppose that \( B \in \mathcal{B} \), since it is open, then we know that \( B \) intersects \( A \)
Now suppose that every basis element containing \( x \) intersects \( A \), and we'd like to show that \( x \in \bar{A} \) which is equivalent to every open set containing \( x \) intersecting \( A \), so suppose that \( U \in \mathcal{T} \) since \( \mathcal{B} \) generates \( \mathcal{T} \), then there is some \( B \in \mathcal{B} \) such that \( x \in B \subseteq U \) since \( A \) intersects \( B \) then it also intersects \( U \) as needed.