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.

I have a problem:

For $C\left [ 0,1 \right ]=\left \{ x:\left [ 0,1 \right ] \to \Bbb R \ \text{is continuous on } \left [ 0,1 \right ] \right \}$, with a norm: $$\left \| x \right \|=\sup_{t\in \left [ 0,1 \right ]}\left \{ \left | x(t) \right | \right \}$$ Let $A=\left \{ x \in C\left [ 0,1 \right ]:\ 0=x(0)\le x(t) \le x(1)=1,\ \forall t\in \left [ 0,1 \right ] \right \}$.

Show that $A$ is non-compact.


I have thought about my problem and I'll write down what I have tried:

Every sequence in $A$ has a convergent subsequence, whose limit lies in $A$.

Therefore, I write:

$$ \text{A is compact} \iff \forall \left \{ x_n \right \}\subset A, \ \text{then} \ \exists \left \{ x_{n_k} \right \}\subset \left \{ x_n \right \}: x_{n_k} \to a \in A $$

I take $\left \{ x_n \right \}=t^n,\ \forall t\in \left [ 0,1 \right ]$ then $ \left \{ x_n \right \}\subset A$, because

  • $ x_n \in C[0,1]$;

  • $0=x_n(0)\le x_n(t) \le x_n(1)=1,\ \forall t\in \left [ 0,1 \right ] $.

But I have trouble when I try to show that $$\not\exists \left \{ x_{n_k} \right \}\subset \left \{ x_n \right \}: x_{n_k} \to a \in A $$

Any help will be appreciated. Thanks!

share|improve this question

1 Answer 1

up vote 1 down vote accepted

We have $x_n(t) \to 0$ if $t \in [0,1)$, and $x_n(1) \to 1$. Hence $x_n$ cannot have any convergent subsequence.

If $x_{n_k} \to x$, then $x$ would be continuous since uniformly convergent, and $x_n(t) \to x(t)$ for all $t \in [0,1]$.

share|improve this answer
Yes! Your explanations are easy to understand for me. Now, I see :) . Thanks copper.hat. –  kimtahe6 Jan 21 '14 at 7:23
You are very welcome. –  copper.hat Jan 21 '14 at 7:26

Your Answer


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.