How to finish this proof about compact implies bounded by the open cover definition

Question

If $K$ is compact then it is bounded. (that it is closed was very easy to prove)

Definitions

.A set is called compact if every sequence has a convergent subsequence.

Attempt

Let $K$ be compact and assume that it is not bounded. Let $x \in K$ be any point. Then for every $n \in \mathbb N$ let $x_n$ be such that $d(x,x_n) > n$.

Now I want to show that $x_n$ does not have a convergent subsequence. I tried by contradiction: assume $x_{n_k}$ was a convergent subsequence and $x_{n_k}\to y$ for some $y \in K$. How to proceed?

Answer 1

To complete the proof indicated (although I Balla's approach is generally better) show that $x_k$ is not cauchy, given $n_0$ for any $m>n_0$ we have

$$d(x, x_{m}) \leq d(x, x_{n_0}) +d(x_m, x_{n_0})$$ and so

$$d(x, x_{m}) -d(x, x_{n_0}) \leq d(x_m, x_{n_0})$$ and by chosing $m$ large enough we have that $d(x_m, x_{n_0})$ is large.

Answer 2

Hint: Given $x_1, \ldots, x_n$, since $K$ is unbounded, $\displaystyle\exists x_{n+1} \in K \setminus \bigcup_{i=1}^nB(x_i, 1)$. Then $d(x_i, x_{n+1}) \ge 1\ \forall i$.

Answer 3

Alternatively you can use the equivalently definition that :

$K$ is compact $\iff $ Every open cover of K has a finite subcover

Let $U_1(x)$ be a ball with radius $1$ around x. Then cover your set with those balls and use the compactness to get a finite cover of those balls with radius 1. Its easy to conclude now that your set is bounded above.

Edit: (Second Proof)

Let's assume that K is unbounded. We now want to construct a sequence which does not have a convergent subsequence in K.

Let $x_0\in K$ be arbitrary. We choose $x_1\in K$ such that $d(x_1,x_0)\geq1$. We will define the next parts of the sequence recursively:

If we already got $x_0,x_1,....,x_k$ then $R_k$ is the radius of a circle about $x_0$ which contains $x_1,....,x_k$. We choose now $x_{k+1}$ such that $d(x_{k+1},x_0)\geq R_k+1$. For all $l \leq k$ we conclude:

$d(x_{k+1},x_l)\geq d(x_{k+1},x_0)-d(x_0,x_l)\geq R_k+1-R_k=1 $

So, any two parts of $(x_n)_{n\in \mathbb N}$ have a distance at least 1 to each other, hence this sequence does not contain a convergent subsequence.

Answer 4

To keep in touch with your previous work,

if you suppose that $x_{n_k}\to y$, note that the function $h\to d(x,h)$ is continuous.

Therefore, the sequence $d(x,x_{n_k})_k$ converges to $d(x,y)$.

But since $d(x,x_{n_k}) > n_k$ and the sequence $(n_k)_k$ is increasing, the sequence $d(x,x_{n_k})$ is unbounded and convergent.

Contradiction.

---

A different approach

Suppose for contradiction that $K$ is unbounded.

Negating the definition of boundedness, one derives the existence of two sequences $(x_n)$, $(y_n)$ such that $\forall n\in \mathbb N, d(x_n,y_n)>n$.

Since $K$ is compact, there is a convergent subsequence for $(x_n)$, say $\large(x_{n_k})$.

The first trick: since $K$ is compact, the sequence $\large(y_{n_k})$ has a convergent subsequence, say $\large(y_{n'_j})$ (I changed notations for clarity).

The second trick: the sequence $\large(x_{n'_j})$ is convergent since it is a subsequence of $\large (x_{n_k})$.

Therefore, the sequence $(\large d(x_{n'_j},y_{n'_j}))$ converges, and is unbounded.

Contradiction.