Consider the topology on $[-1, 1] \times [-1, 1] \subset \mathbb R^2$ where the set is considered open if and only if it is empty, or contains $(0,0)$. Prove: This definition of open sets defines a valid topology. Determine whether the space is Hausdorff, and prove your claim.
Solution: Let $\tau = \{U \subset [-1, 1] \times [-1, 1] : U = \emptyset \ or \ (0,0) \in U \}$. $\tau $ is a topology:
- $\emptyset \in \tau$
- $[-1, 1] \times [-1, 1]$ contains $(0,0)$ so $X = [-1, 1] \times [-1, 1] \in \tau$
- Let $ \{U_i \}, i \in I$ be a collection of elements of $\tau$. Then for all $U_i = \emptyset$, thus union of $U_i, i\in I = \emptyset$ so $\bigcup_{i \in I} (U_i) \in \tau$.
- Let $ \{U_i \}, i \in A, A$ is finite, be a collection of elements of $\tau$. If there exists $U_i$ with $(0,0) \notin U_i$ then this $U_i = \emptyset$ so $\bigcap_{i \in A} (U_i) = \emptyset$. Otherwise $(0,0) \in U_i$ for all $i \in A$. So $(0,0) \in \bigcap_{i \in A} (U_i)$. Hence the finite intersection of $U_i \in \tau$.
The topology $\tau$ is not Hausdorff. For suppose $\tau$ is Hausdorff. Since $(-1,0) \neq (1,0) $ there would be open sets $U, V$ such that $(-1, 0) \in U$, $(1,0) \in V$ that are disjoint. However since $U \neq \emptyset$ and $V \neq \emptyset$, and $(0,0) \in U$, $(0,0) \in V$ So $(0,0) \in U \bigcap V$. Hence $U$ and $V$ cannot be disjoint.