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Consider the space of continuous, real-valued functions on $[0,1]$ under the sup-norm metric. I believe that the set of functions {$x^n$} for all natural n is a closed and bounded subset of our metric space that is not compact. It is bounded as the difference between any two functions in the set is at most $1$ and I believe it is not compact as it does not contain a convergent subsequence (no subsequence converges uniformly to a continuous function, let alone a continuous function in the set). I don't have a rigorous proof of closure but do believe it is closed. Hence, I believe that it works as an example of a subset of a metric space that is closed and bounded but not compact. Is my reasoning valid?

2 Answers 2

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Suppose it is not closed. Its complementary $U$ is not open. There exists $f\in U$ such that every ball $B(f,1/n)$ contains a function $x^m$. We deduce that $f=lim_nx^{u(n)}$. Since $f$ is not of the form $x^m$. $lim_nu(n)=+\infty$. But In this case, $lim_nx^{u(n)}$ is the function $g:[0,1]\rightarrow R$ such that $g(x)=0, x\neq 0$ and $g(1)=1$. Contradiction since $g$ is not continue, and thus cannot be equal to $f$.

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    What do you mean by $u(n)$? But yes I do agree that it is closed. So does the rest of the argument work? Is it a closed and bounded set that is not compact? – 2017-01-01
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    u(n) is a sequence of integers, such that the functions $x^{u(n))}\in B(f,1/n)$. That is to clarify the fact that it is closed since in your question, you say that you don't have a rigorous proof of that. – 2017-01-01
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    Right ok, I understand and it is pretty much the same argument that I use to show that it is not compact. – 2017-01-01
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    I am still curious as to whether that argument also shows that it is not compact, though. – 2017-01-01
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    This argument shows also that it s not compact, since the sequence of functions $x^n$ does not have a converging subsequence since such a sequence will converges towards $g$ such that $g(x)=0, x\neq 1$ and $g(1)=1$ which is not continue. – 2017-01-01
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    Ok yes, that is what I thought, thank you! – 2017-01-01
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Assume that $\,f_n(x)=x^{k_n}\to f(x),\,$ is uniformly convergent in $[0,1]$, where $\,f\in C[0,1]$. Then, for every $a\in (0,1)$, the $\{\,f_n\}$ shall be uniformly converging in $[0,a]$. But, $x^n\to 0$, uniformly convergent in $[0,a]$, since $$ \max_{x\in[0,a]}\|\,x^n\|=a^n\to 0,\quad\text{as $n\to\infty$.} $$ Hence $\{\,f_n\}$ is also uniformly converging in $[0,a]$, and $\,f|_{[0,a]}=0$. Thus $\,f(x)=0$, for all $x<1$, and as $\,f$ is continuous, then $f\equiv =0$.

But, $\,f_n(1)=1^{k_n}=1$, and thus, $f(1)=\lim_{n\to\infty}f_n(1)=1$. Contradiction.

Thus $\{x^n\}$ does NOT have any accumulation point, and hence it is closed.

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    Right ok, that makes sense too, thank you! And you agree that since there is no converging subsequence it is not compact, right? – 2017-01-01
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    To Vr Jay : Yes. In a metric space any sequence in a compact subset has a sub-sequence converging to a point in that set. – 2017-01-02