Is this space $T_1$ and satisfies property $P$?.

Question

Question: let $X=\{1,2,3\}$ and $τ=\{\{1\},\{2\},\{1,2\},\{2,3\},\{1,2,3\}\}$. The topological space $(X,τ)$ is said to have property $P$ if for any two proper disjoint closed subsets $Y$ and $Z$ of $X$, there exists disjoint open sets $U$, $V$ such that $Y ⊆U, Z⊆V$. Then the topological space $(X,τ)$

(A) is $T_1$ and satisfies $P$

(B) is $T_1$ but does not satisfies $P$

(C) is not $T_1$ and satisfies $P$

(D) is not $T_1$ and not satisfies $P$

I am beginner of topology, and I don't have key of above problem , so please help me....

My attempt: what I know is , members of $τ$ are called open sets and I know $(X,τ)$ will be $τ_1$ if Singleton's are closed set, but there is no set in $τ$ such that whose complement is $\{2\}$ and hence $\{2\}$ is not closed set in $(X,τ)$ and hence space is not $τ_1$.

Now to see whether $(X,τ)$ satisfies $P$. Clearly I can see, $\{1\}$ and $\{2, 3\}$ are closed sets(because they are complement of open sets) and they are disjoint too and clearly there exists open sets $U, V$ such that $\{1\}$ and $\{2, 3\}$ are contained in $U$ and $V$ respectively(here $U$, $V$ are $\{1\}$ and $\{2, 3\}$ resp.) Hence $(X,τ)$ satisfies $P$.

Please check whether "my attempt" is correct or not! and is above solution by me is complete and correct? as i beginner and don't have key, I don't know I am correct or not. please help me.

Answer 1

First of all, remember to add $\emptyset$ as open subset.

Your proof of $X$ not being $T_1$ is correct.

But, to prove that $X$ satisfies $P$, you have to show that every pair of disjoint closed subsets have disjoint open neighborhoods containing them. What you did is finding one pair with that property. So that part of the proof is not correct.

Answer 2

Your space is not $T_1$ since $$\{2\}^c = \{1,3\} \notin \tau$$

so $\{2\}$ is not a closed set.

All closed sets in your space are

$$\{\{2,3\}, \{1, 3\}, \{3\}, \{1\}, \{1, 2, 3\}, \emptyset\}$$

The only nontrivial disjoint pairs are $(\{2,3\}, \{1\})$ and $(\{3\}, \{1\})$ both of which can be separated by $\{2, 3\}, \{1\} \in \tau$.

Therefore, your space indeed satisfies $P$ so the correct option is $(C)$.

Answer 3

Your proof is correct but not complete.

You just found one example of two closed disjoint sets and their corresponding disjoint open sets. Apparently there are other disjoint closed sets in your topology.

Try to find all of them.