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 ______.
Show 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.
Step 1: Notice the partition structure. $X = \{5,6,7,8,9,10\}$ splits into two disjoint blocks, $B_1 = \{5,6,7\}$ and $B_2 = \{8,9,10\}$, with $B_1 \cup B_2 = X$. The topology $\tau$ is built only from unions of these two blocks: the empty union $\phi$, $B_1$ alone, $B_2$ alone, and the full union $X$. So $\tau$ has exactly 4 members and no others.
Step 2: A subset can be open only if it equals one of these 4 sets. $\tau$ lists every open set directly, so being open just means being one of $\{\phi, B_1, B_2, X\}$. Any other subset of $X$ is, by definition, not open.
Step 3: A subset can be closed only if its complement is one of these 4 sets. Take complements of $\phi, B_1, B_2, X$ inside $X$: the complement of $\phi$ is $X$, of $X$ is $\phi$, of $B_1$ is $B_2$, and of $B_2$ is $B_1$. So the complements of the 4 open sets are again exactly $\{X, \phi, B_2, B_1\}$, the same 4 sets. This means the closed sets are also precisely $\{\phi, B_1, B_2, X\}$, nothing more.
Step 4: Combine and count. Since the open sets and the closed sets are literally the same 4 sets, the total number of subsets that qualify as open, closed, or both is just 4. The total number of subsets of a 6 element set is $2^6 = 64$. Every subset outside these 4 special ones fails both tests, so it is neither open nor closed.
\[ 64 - 4 = 60 \]
