Determine the closure of the set $K=\{\frac{1}{n}\mid n\in\mathbb N\}$ under each of topologies The questions are the following:
Consider the five topologies on the real line $\mathbb R$:


*

*$\mathcal T_1$: the standard topology

*$\mathcal T_2$: the $K$-topology

*$\mathcal T_3$: the finite complement topology

*$\mathcal T_4$: the upper limit topology

*$\mathcal T_5$: the topology generated by the basis $\{(-\infty,a)\mid a \in \mathbb R\}$
Determine the closure of the set $K=\{\frac{1}{n}\mid n\in\mathbb N\}$ under each of these topologies.
My answer is the following:
$\mathcal T_1$: $\mathrm{cl}(K)=\{0\} \cup K$.
$\mathcal T_4$: $\{0\} \cup K$.
$\mathcal T_5$: $[0, \infty)$
Thank you.
 A: $\mathcal{T}_1:$ That's right, well done. Note that this topology is generated by the intervals  $(a,b)$. So for any point $x$ in $\mathbb{R}$ in between $k = \frac1n$ and $k^\prime = \frac1m$ we can choose $(x - m, x + m)$ where $m = \frac12 \min (\frac1n , \frac1m)$ to see that $x$ is not in the closure of $K$. Looking at $0$ we see that every interval $0 \in (a,b)$ has to have $0 < b$ so there will be an $n$ such that $0 < \frac1n < b$ and hence every open set containing $0$ will have non-empty intersection with $K$. Hence we get $\overline{K} = K \cup \{0\}$.
$\mathcal{T}_2$: Note that $\mathcal{T}_2$  is generated by $(a,b) $ and $ (a,b) \setminus K $ so $\mathcal{T}_1 \subset \mathcal{T}_2$ and hence $ \overline{K}_{\mathcal{T}_2} \subset \overline{K}_{\mathcal{T}_1}$ since we have more sets that might contain our point but might not intersect with $K$. Since $K \subset \overline{K}$ we therefore know that the only point we need to think about is $0$. But none of the sets we added to the basis intersect with $K$ hence there now are open sets $O$ containing $0$ with $O \cap K = \varnothing$. Hence $0$ is not in the closure of $K$ and we have $\overline{K}=K$.
$\mathcal{T}_3:$ Note that the open sets in this topology look like this: $O = \mathbb{R} \setminus \{x_1, \dots, x_n \}$. Consider a point $p$ in $\mathbb{R}$. Let $O$ be an open set containing $p$. Note that $O$ has non-empty intersection with $K$ since $O$ contains all points of $\mathbb{R}$ except finitely many. In particular, $K$ contains infinitely many points hence their intersection will be non-empty. Since $p$ was arbitrary we get $\overline{K} = \mathbb{R}$.
$\mathcal{T}_4$: Note that the upper limit topology is generated by intervals of the form $(a, b]$. Note that for a point $p \neq 0$ outside $K$ we can pick $(\varepsilon , p]$ with $\varepsilon$ small enough such that $(\varepsilon , p] \cap K = \varnothing$. For $0$ we pick $(a, 0]$ to get a set that does not intersect with $K$ hence $0$ is not in the closure of $K$. Hence we get $\overline{K} = K$.
$\mathcal{T}_5$: Again you have it right, well done. To see this pick a point $p$ to the right of $0$ or $0$ itself and note that every open interval $(-\infty, a)$ containing $p$ will intersect with $K$ since there will be an $n$ such that $\frac1n < p$. 
Hope this helps.
A: You just use the definition of the closure, for example: $x \in cl(K)$ if and only if every open neighborhood $V$ containing $x$ intersects $K$. Use the obvious $K \subset cl(K)$. Then think of any possible limit points by looking at the form of the neighborhoods.
