2
$\begingroup$

True or false: Let $X \subseteq \Bbb Q^2$. Suppose each continuous function $f:X \to \Bbb R^2$ is bounded. Then $X$ is finite.

Now it will be compact for sure just by using distance function.

Now what can we do?

$\endgroup$
6
$\begingroup$

Hint: Consider $X = \{(1,0), (1/2,0), (1/3,0), ..., (0,0)\}$. What can we say about the behaviour of $f(x)$ as $x\to (0,0)$?

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ $f(x) \to f(0,0)$, then? $\endgroup$ – Ri-Li Jul 13 '19 at 1:18
  • 1
    $\begingroup$ Yes. There is an important property that any convergent sequence must have, which is ... ? $\endgroup$ – BigbearZzz Jul 13 '19 at 1:22
  • $\begingroup$ A compact set will contain the limit point. And convergent sequence must be bounded. Is there any other thing? $\endgroup$ – Ri-Li Jul 13 '19 at 1:44
  • $\begingroup$ Oh ok. So it has to be bounded and that is why the statement is false. Am I right? $\endgroup$ – Ri-Li Jul 13 '19 at 1:45
  • $\begingroup$ You are right. For more sophisticated answers you can see the other two here. $\endgroup$ – BigbearZzz Jul 13 '19 at 8:36
1
$\begingroup$

In a metric space $(X,d)$, a subspace $Y \subseteq X$ has the property that all continuous $f: Y \to \Bbb R$ are bounded ($Y$ is then called pseudocompact) iff $Y$ is a compact subspace of $X$. See here e.g. And $X=\mathbb{Q}^2$ has many infinite compact subsets, like any convergent sequence together with its limit.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

Any subspace $X\subset \Bbb R^2$ is compact iff $X$ is closed in $\Bbb R^2$ and bounded with respect to the usual metric.

A continuous image of a compact space is compact, so if $X$ is compact and $f:X\to \Bbb R^2$ is continuous then the image $f(X)$ is compact, hence $f(X)$ is bounded.

For example $X=\{(0,0)\}\cup \{(1/n,0):n\in \Bbb N\}$ is closed and bounded.

Remarks. (i). If $X$ is an unbounded subset of $\Bbb R^2$ then id$_X:X\to \Bbb R^2$ is continuous and has unbounded range. (ii). If $X$ is a non-closed subset of $\Bbb R^2,$ take $p\in \overline X\setminus X$... And with the usual distance-function (metric) $d$, let $f(x)= (1/d(x,p),\,0)$ for $x\in X $. Then $f:X\to \Bbb R^2$ is continuous and unbounded.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.