Compactness of function space Let $X$ be the set of all continuous functions from the unit interval $[0,  1]$ to itself.
(1) Prove that $X$ is not compact under the pointwise convergence topology.
(2) Prove that $X$ is compact under the uniform convergence topology.
To prove (1), I found the counterexample as the sequence of functions $\langle f_n \rangle$ defined as $f_n (x) = x^n$, which converges to $f(x) = 0$ for $0 \le x <1$ and $f(x)=1$ for $x=1$, which is not continuous, so $X$ is not sequentially closed. However, I remember that it cannot be guaranteed that sequential closedness does not imply closedness of topological space unless it is first countable, and $I^I$ under product topology is not first countable. How should I verify that $X$ is not compact?
Also, this counterexample can be applied in (2), which deduces that $X$ is not compact under the uniform convergence topology. Is there anything wrong in my argument? 
As I know, in the metric space, uniform convergence topology is the induced topology from the uniform metric $d$ between two functions is given as $d(f, g) = \sup|f-g|$, right? Also, in the metric space, sequential compactness and compactness are equivalent. Is there any error in what I know?
 A: In the pointwise case, we do know that compactness implies limit point compactness in all spaces (metrisable or not). So if the space were compact there would be a limit point $f$ of $\{f_n: n \in \mathbb{N}\}$. As projections (point-evaluations) are continuous, they preserve limit points, so $f(x)$ is a limit point of $\{f_n(x): n \in \mathbb{N}\}$ for all $x \in [0,1]$. For all $x$ this limit point is unique, so this forces $f(x) = 0$ for $x <1$ and $f(1) =1$. So this yields the contradiction as $f$ is a limit point in the compact set $[0,1]^{[0,1]}$ but is not in its subspace $C([0,1], [0,1])$. So the set $\{f_n: n \in \mathbb{N}\}$ has no limit point in the space at hand, so this space is not compact. 
For the sup-metric, sequential compactness is indeed necessary and sufficient, so there you might need the classical fact that if $f_n \rightarrow f$ uniformly on $[0,1]$ and all $f_n$ are continuous then so is $f$. Your $f_n$ do not converge to $f$ in the sup-metric, it follows from this. 
Then as the identity $i$ is continuous from $C([0,1], [0,1]), \mathcal{T}_{\sup})$ to $C([0,1], [0,1]) ,\mathcal{T}_{\text{pw}})$, as the topoloy is weaker, the left hand space cannot be compact either, or the first would have been. I think the problem as stated is false. Both are non-compact.
A: By Tychonoff, $I^I$ is compact in the product topology, which is the same as the topology of pointwise convergence. It is clearly Hausdorff, so it suffices to show $C[0,1]\subset I^I$ is not closed. For this, consider $f:I\to I$ defined by 
$f(x)= \begin{cases} 
      0 & 0\le x\leq 1/2 \\
      1 & 1/2< x \le 1
   \end{cases}
$
and observe that any (finite) intersection of subbasis elements $S(x,U)\ni f$ contains a continuous function. 
For the second item, I think your own example works to show $C(I,I)$ is $not$ $compact$ in the sup norm toplogy which is the same as that of compact convergence, because $I$ is a metric space:
$f_n(x)=x^n$ converges $pointwise$ to 
$f(x)= \begin{cases} 
      0 & 0\le x<1 \\
      1 & x=1
   \end{cases}
$ 
Thus, any subsequence of $f_n$ that converged in the sup norm topology would have to converge to $f$ as well. This argument shows that no subsequence of $f_n$ can converge to a continuous function in the sup norm topology, and therefore the space is not sequentially compact.
A: On compactness and completeness of $I^I$ and $C(I,I)$
Ok, I know this is already answered, but, working on it, I clarified myself some points I would like to share.
Let $X=Y=I$, where $I=[0,1]$. Let $\cal{D}_Y$ the unique uniformity of the compact $Y$ giving its topology. The space $I^I$ is compact in the pointwise topology $\cal{T}_p$ (product topology), so it has a unique uniformity $\cal{D}_p$ giving its topology. The basis for this uniformity is consisted of the sets $E_D=\{(f,g):(f(x),g(x))\in D$, for every $x \in F, F$ finite set in $X \}$ , $D\in \cal{D}_Y$. $(I^I, \cal{T}_p)$ is not a metric space though, since $X=I$ is uncountable.
However, on $I^I$ we may consider also the topology of uniform convergence $\cal{T}_u$ which is stronger than $\cal{T}_p$ and it is derived by the uniformity of uniform convergence $\cal{D}_u$ with basis consisting of all the sets of the form $E_D=\{(f,g):(f(x),g(x))\in D, \forall  x\in X\}$, $D\in \cal{D}_Y$. This topology coincides with that derived by the supremum norm $|| . ||_\infty$, so $(I^I, \cal{T}_u)$ is now a metric space, but it is not compact. Sequences are sufficient then to describe its topology.
If $(I^I, \cal{T}_u)$ was compact, then its closed subset $C(I,I)$ of continuous functions would be compact too. This is not the case, as it has already mentioned in the previous answers and comments.  $C(I,I)$ is closed in this topology, since if $\{f_n\}$ is a sequence in $I^I$ with $f_n \rightarrow f \in I^I$ in sup norm, then $f$ is continuous too. So, $C(I,I)$ is a complete subspace of the complete space $(I^I, \cal{T}_u)$.
