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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?

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    $\begingroup$ You are using the definition of sequentially compact, which is equivalent to compactness for metric spaces. However it's perhaps easier in this case to apply the compact definition directly by exhibiting an open cover for an unbounded set with no finite subcover. $\endgroup$
    – hardmath
    Jul 20, 2014 at 15:20
  • $\begingroup$ This proof requires AC, doesn't it? $\endgroup$ Jul 14, 2018 at 10:18

4 Answers 4

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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.

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  • $\begingroup$ No I know how to prove it using that definition. My question how to prove it using the other definition. $\endgroup$
    – blue
    Jul 20, 2014 at 17:44
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    $\begingroup$ Added an alternatively prove for you. $\endgroup$
    – Marm
    Jul 22, 2014 at 0:56
  • $\begingroup$ What a simple proof! Thanks $\endgroup$ Mar 19 at 1:18
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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.

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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.

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  • $\begingroup$ Why is $d(x,x_m)-d(x,x_{n_0})$ large? $\endgroup$
    – blue
    Jul 20, 2014 at 17:51
  • $\begingroup$ @blue $d(x,x_m)>m$, and $d(x,x_{n_0})$ is fixed $\endgroup$ Jul 20, 2014 at 17:58
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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 $d(x_i, x_{n+1}) \ge 1\ \forall i$.

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  • $\begingroup$ Does it mean I have to construct a different sequence and cant use the sequence in my question? $\endgroup$
    – blue
    Jul 20, 2014 at 17:46

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