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!
 A: It's obviously that $0$ is the only accumulation point of that set. And we can see that $0 \notin T$. So $T$ is not closed(closed set is the set that includes all its accumulation points).
A: 
A subset $A \subseteq \Bbb{R}$ is closed if and only if it contain all of its limit points.

With standard topology on $\Bbb{R}$, the subset $T$ is not closed, since it has a limit point $0$ which is not in $T$. Alternatively, you can show that its complement is not open : any neighbourhood of $0 \in \Bbb{R}\smallsetminus  T$ contain infinitely many elements of $T$. So $0$ doesn't have a neighbourhood contained in $\Bbb{R}\smallsetminus T$. Hence $\Bbb{R}\smallsetminus T$ is not open.
A: Your answer does not work because $\mathbb{R}\setminus T$ does not contain the interval $(-\infty, 1)$. For example, $\frac12\in T$.
One way to solve your question: can you show that if $T$ is closed, you must have $0\in T$? 
A: 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.
A: It turns out that$$\mathbb{R}\setminus T=\left(\bigcup_{n=2}^\infty\left(\frac1{n-1},\frac1n\right)\right)\cup(-\infty,0]\cup(1,+\infty),$$which is not an open set.
