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

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?

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

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

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.

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.