Question

Let \( X = \{5, 6, 7, 8, 9, 10\} \) be equipped with the topology
\[ \tau = \{ \phi, X, \{5,6,7\}, \{8,9,10\} \} \]
Then the number of subsets of \( X \) which are neither open nor closed is ______.

Hint:

Find the closed sets as complements of the open sets first, then subtract the open-or-closed count from 2 to the power 6.

Answer 1

Correct Answer: 60

Step 1: Understanding the Question.
The set \( X \) has 6 elements, so it has \( 2^6 = 64 \) subsets in total. A subset is called open if it belongs to \( \tau \), and closed if its complement (in \( X \)) belongs to \( \tau \). We need to count the subsets that fail both tests, that is, subsets that are neither open nor closed.

Step 2: List the open sets.
By definition, the open sets are exactly the members of \( \tau \):
\[ \phi, \quad X, \quad \{5,6,7\}, \quad \{8,9,10\} \]
That gives 4 open sets.

Step 3: Find the closed sets by taking complements.
A set is closed when \( X \) minus that set lies in \( \tau \). Take the complement of each open set:
\[ X \setminus \phi = X, \qquad X \setminus X = \phi \]
\[ X \setminus \{5,6,7\} = \{8,9,10\}, \qquad X \setminus \{8,9,10\} = \{5,6,7\} \]
All four of these complements, \( X, \phi, \{8,9,10\}, \{5,6,7\} \), are themselves already in \( \tau \), so every open set here is also closed. The closed sets are the same four sets: \( \phi, X, \{5,6,7\}, \{8,9,10\} \).

Step 4: Count the sets that are open or closed.
Because the list of closed sets is identical to the list of open sets, the union of open sets and closed sets still has only 4 sets. No new set gets added by including the closed ones.

Final Answer:
Subsets that are neither open nor closed equal total subsets minus open-or-closed subsets:
\[ 64 - 4 = 60 \]
\[ \boxed{60} \]