# Show that $A$ is non-compact

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.

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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!

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## 1 Answer

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 $t \in [0,1]$.

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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