Exercise on Limit Points in Topology 2
In this exercise, I need to determine the set of limit points of \( A = \{a, c\} \) in the space \( X = \{a, b, c\} \) with the topology \(\{X, \emptyset, \{a\}, \{a, b\}\}\).
A point \( x \) is a limit point of a set \( A \) if every neighborhood of \( x \) contains at least one point of \( A \) other than \( x \).
The topology on \( X = \{a, b, c\} \) is \(\{X, \emptyset, \{a\}, \{a, b\}\}\). Thus, the open sets are:
- \( \emptyset \)
- \( \{a\} \)
- \( \{a, b\} \)
- \( X \) (i.e., \( \{a, b, c\} \))
To identify the limit points of \( A = \{a, c\} \), I need to examine each point \( x \) in the space \( X = \{a, b, c\} \) and verify whether it contains at least one point of \( A \) other than \( x \).
- Point \( a \)
The neighborhoods of \( a \) are \(\{a\}\), \(\{a, b\}\), and \( X \).- \(\{a\}\) contains only \( a \) and no other points of \( A \). Since \( x = a \), this does not satisfy the condition as the neighborhood must include at least one point of \( A \) other than \( x \).
- \(\{a, b\}\) does not contain \( c \), so it does not meet the condition.
- \( X \) contains \( c \), making it the only neighborhood that meets the condition. However, for \( a \) to be a limit point, every neighborhood of \( a \) must contain a point of \( A \) other than \( a \).
- Point \( b \)
The neighborhoods of \( b \) are \(\{a, b\}\) and \( X \).- \(\{a, b\}\) contains \( a \), which is in \( A \). Therefore, it meets the condition.
- \( X \) contains both \( a \) and \( c \), both of which are in \( A \). This neighborhood also meets the condition.
- Point \( c \)
The point \( c \) has only one neighborhood: \( X \).- \( X \) contains \( a \), which is in \( A \). Therefore, it meets the condition.
The limit points of \( A = \{a, c\} \) in the space \( X = \{a, b, c\} \) with the topology \(\{X, \emptyset, \{a\}, \{a, b\}\}\) are \( b \) and \( c \).
Thus, the set of limit points of \( A \) is \(\{b, c\}\).
