Is the set $T = \{1,\frac12,\frac13,\ldots,\frac1n,\ldots\}$ closed

general-topologyreal-analysis

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!

Best Answer

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).

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