Is set $T$ is complete

Question

let let $$S = \{\frac{1}{n}: n \in \mathbb{N} \}\cup\{0\}$$ and $$T= \{n+ \frac{1}{n}: n\in\mathbb{N} \}$$ be the subset of the metric space $\mathbb R$ with usual metric, then,

A) $S$ is complete but not $T$

B) $T$ is complete but not $S$

C) Both $T$ and $S$ are complete

I know that $S$ is complete because here $0$ is contained in set $S$ as limit point is contain in the set $S$......but $0$ is not contained in set $T$ so $T$ will not complete ,,as option 1 is the correct answer that is $S$ is complete but not $T$....

pliz verified am I right or wrong? thanks in advance

Answer 1

The option C) is correct. The set $T$ is complete because $T$ is a closed subset of $\mathbb R$ and $\mathbb R$ is complete.

Answer 2

It is easy to see that $$S = \{\frac{1}{n}: n ∈ \mathbb{N} \}\cup\{0\}$$ is a closed subset of $\Bbb R$ since $1/n\to 0.$

On the other hand, if $x_j = n_j+\frac{1}{n_j} \in T$ is a sequence converging to $x\in \Bbb R$

Then, There exists $j_0$ such that for $j>j_0$ we have,

$$|x-n_j|\le \frac{1}{n_j}+\left|x-n_j-\frac{1}{n_j}\right| <\frac{1}{n_j}+\frac12<\frac{3}{2} $$

That is for $j>j_0$ we have, $n_j\in (x-\frac{3}{2}, x+\frac{3}{2})$ but the interval only finite number of integers. Hence the sequence $n_j$ must be stationary. Namely there exists $k\in\Bbb N$ such that $$x= n_k+\frac{1}{n_k}\in T.$$ hence T will be closed therefore complete.