Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

To prove:

"If every continuous function $f$ in $X \subset \mathbb{R}^2$ is bounded then $X$ is compact."

My attempt :

In $\mathbb{R}^n$ a set $X$ is compact iff it is closed and bounded. I can show $X$ is bounded, but can not prove $X$ is closed.

The distance function $f : X \rightarrow \mathbb{R}$ defined by $f(x , y) = \sqrt{x^2 + y^2}$ is continuous on $X$. So it is bounded. Thus distance between any two points of $X$ is finite and hence $X$ is bounded.

Let $a \in \mathbb{R}^2$ be a limit point of $X$. Consider a sequence $\{a_n\}$ in $X$ converges to $a$. So for any continuous function $f$, the sequence $\{f(a_n)\}$ converges to $f(a)$ and $f(a)$ is finite. It does not imply $a \in X$.

Thank you for your help.

share|improve this question
    
Mr.Prahlad's answer for my question math.stackexchange.com/questions/577878/… might be helpful for you I believe. –  Praphulla Koushik Nov 23 '13 at 6:17
2  
@Samprity : To show that $X$ is bounded, by definition you must show that $X \subseteq \left\{ (x,y) \in \mathbb R^2 \, \left| \, \sqrt{(x-x_0)^2 + (y-y_0)^2} < R \right. \right\}$ for a certain $R > 0$ and $(x_0,y_0)$ (this set is often abbreviated $B_R(x_0,y_0)$. Your argument precisely shows that $X$ is a subset of such a set for $x_0=y_0=0$ and $R$ large enough. Logically speaking, "Thus distance between any two points of $X$ is finite" is a consequence of the boundedness of $X$, not the other way around. :) –  Patrick Da Silva Nov 23 '13 at 6:33
    
@PraphullaKoushik Yes Mr. Vidyanathan's answer is helpful. –  Dutta Nov 23 '13 at 6:55

1 Answer 1

up vote 2 down vote accepted

To prove that $X$ is closed, if $(x_0,y_0)$ is a limit point of $X$ that is not in $X$, consider the function $(x,y) \mapsto 1/f(x-x_0,y-y_0)$; this is just the reciprocal of the distance between the points $(x,y)$ and $(x_0,y_0)$.

share|improve this answer
    
His $f$ doesn't mean the metric on $\mathbb R^2$, but the "distance function on $\mathbb R^2$", i.e. the norm. You got confused. Very unusual to call "$f$" a metric, but your argument works for the closed part. –  Patrick Da Silva Nov 23 '13 at 6:30
1  
@PatrickDaSilva Oops, I'll fix it. Thanks. –  Trevor Wilson Nov 23 '13 at 6:31
    
Well now your $f$ is supposed to be his $f$, it's not the metric! I corrected it. –  Patrick Da Silva Nov 23 '13 at 6:34
1  
@PatrickDaSilva D'oh! Thanks for catching that. –  Trevor Wilson Nov 23 '13 at 6:36
    
@ Mr. Wilson. Thank you for you hints. When $y \rightarrow x$, the function $\frac{1}{f(x,y)} \rightarrow \infty$. If $f(x,y) \neq 0$, $\frac{1}{f(x,y)}$ is continuous, that leads us to a contradiction. So $x$ must be in $X$. Hope it is right and the matter is clear to me. –  Dutta Nov 23 '13 at 6:36

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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