$X=\mathbb{R}$, $\tau_{0}=\{U \subseteq \mathbb{R}|U \subseteq \mathbb{R} \;\text{open in the standard topology and } 0 \in U\} $. Find the limit points of the sequence $x_n=1+\frac{1}{n}$.
Solution:
The sequence $(x_n)$ is a convergent sequence in $(\mathbb{R},\tau_{0})$ and: $0$, and $n\ge1$ are the limit points of $x_n$.
Defintion: A sequence $(x_n)_{n\in \mathbb N}$ in a topological space $X$ is called convergent if $\exists\;x\in X$ such that given any neighborhod $U\subseteq X$ of $x$, $\exists$ $n_U$ s.t. whenever $n \ge n_U$, $x_n\in U$.
Now, using the definition above and defintion of open sets in $\tau_0$, clearly any neighborhood of $1$ contains $0$ and so $0,1$ are limit points of $x_n$.
Now suppose, I choose $k=1.1$, then any neigbhorhood $U_1$ of $1.1$ should contain $0$ (definition of $\tau_{0}$) and choosing $n_U=10$ we have that whenever $n\ge10$, $x_n\in U_1$.
Similarly, I can show that for any real number $n\ge 1$, I can find $n_u$ using archimedian property and hence the limit points are $0$ and any real number $n\ge1$.
Am I correct? Please help if there is any fallacy in my argument. Thanks.
I don't see how $0$ is a limit point as a neighborhood with radius less than 1 would not intersect your sequence. It won't hold for any $n$. For example, take $n = 1$. So if $2$ was a limit point, I could pick radius less than $1/2$ and it will not intersect the sequence aside from the point $2$.