Topological properties of $(0,1)$ with $B=\{ (1/n,1) \}$

Let $X=(0,1);$

$$\tau= \{X,\emptyset,(1/n,1)\mid n\ge 2 \};$$

1)Compact: No.

There is no finite subcover of $X=\bigcup_{n\ge 2}(1/n,1)$.

2)Hausdorff: No.

For $x_1\not =x_2$ there are no disjoint open sets $O,V$ with $x_1\in O$ and $x_2 \in V.$

Assume there are disjoint open sets $x_1 \in O$ and $x_2 \in V.$

Then there are disjoint basis elements $x \in (1/n,1)$, and $x_2 \in (1/m,1)$.

But $(1/n,1) \cap (1/m,1) \not =\emptyset$, for $n, m \ge 2$.

3)Normal: Yes (vacuously). (H.Brandsma's remark)

There are no disjoint closed sets $A,B (\not =\emptyset);$

Assume A, B are disjoint closed sets.

Then $(A\cap B)^c =A^c\cup B^c=X$, where $A^c, B^c$ are open ($c$ for complement),

this implies $A^c=X$ or $B^c=X$.

Hence $A=\emptyset$, or $B=\emptyset.$, a contradiction.

4)Regular: No.

Let $A$ be closed and $x\not \in A$.

Assume there are disjoint open sets $O, U$ s, t.

$A\subset O$ and $x \in U$.

$O=\bigcup_{n\in I, finite} (1/n,1)$, and

$x \in (1/m,1)$ for a $m$.

$O\bigcap U \not =\emptyset$, and we are done.

Kindly check and correct, or give me a hint. Hopefully not completely messed up.

Thank you.

To refute Hausdorff you need two specific points that cannot be separated. So do the work.

If there are no disjoint closed sets then $X$ is normal voidly. ( we cannot refute normality).

For non-regular you need a specific point and disjoint closed set. It helps in the proof that all non-empty open sets of $X$ intersect.