Is the set $T = \{1,\frac12,\frac13,\ldots,\frac1n,\ldots\}$ closed? 
Exercise:Is the set $(T = \{1,\frac12,\frac13,\ldots,\frac1n,\ldots\})$ closed in $\mathbb{R}$?

I tried to answer this question by computing $\mathbb{R}\setminus T=\bigcup_\limits{2}^{\infty}(\frac{1}{n-1},\frac{1}{n})\cup(1,\infty)\cup(-\infty,1)$
So $\mathbb{R}\setminus T$ is the union of open sets hence open which implies $T$ is closed.
The problem arose when I was told that $T$ is not closed.
Question:
What am I doing wrong? How should I solve the question?
Thanks in advance!

By definition of a closed set. Let $M$ be a set. A set $A \subseteq M$ is closed if and only if for any converging sequence $(m)$ where $m_i \in A$, the limit of $m$ lies in the set $M$.
$(\lim_{i \to \infty} m_i) \in A$
In your case, there clearly is a sequence $(m)$ such that $m_i = \frac{1}{n}$. In that case, $\lim_{i \to \infty} m_i = 0 \notin A$. This shows your set is not closed.
