This question was asked in the GATE MA 2023 paper:
Q.44. Let $(\mathbb{R},\tau)$ be a topological space, where the topology $\tau$ is defined as $$\tau = \{U \subset \mathbb{R}: U = \emptyset \ or \ 1 \in U\}.$$ Which of the following statements is/are correct?
- (A) $(\mathbb{R},\tau)$ is 1st-countable
- (B) $(\mathbb{R},\tau)$ is Hausdorff
- (C) $(\mathbb{R},\tau)$ is separable
- (D) The closure of $(1,5)$ is $[1,5]$
I found that it is separable since it has countable dense set, but can't prove it is first countable or not, since 1 and empty set has uncountable neighborhood it implies it has at least countable neighborhood so it is first countable. can I use this logic?
That's the particular point topology of $1$ on $\mathbb{R}$. It is first countable since, for a given $x\in\mathbb{R}$, we know that $\mathcal{B}(x)=\{\{1,x\}\}$ is a neighborhood basis of $x$ (it doesn't matter wether $x=1$ or not). However it isn't Hausdorff since, for every $x\neq1$ and every neighborhood $N_x\in \mathcal{N}(x)$, we have that $\exists A\in\mathcal{T}:A\subseteq N_x\Rightarrow x,1\in A\subseteq N_x$ which means that, for every neighborhood of $1$ $N_1\in\mathcal{N}(1)$, we also have that $1\in N_x\cap N_1\Rightarrow N_x\cap N_1\neq\emptyset$. Finally, since $1$ is not in $(1,5)$ (i.e. its complement contains $1$ and thus is open), we know that it is close which means that its closure is the same $\bf (1,5)$ and not $[1,5]$.