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?


3 Answers 3


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.

  • $\begingroup$ Thank you for the pointwise convergence case. However, for the uniform convergence case, sequential compactness means "any" sequence must have a convergent subsequence, and so does my $f_n$, but it seems that the $f_n$ does not have any convergent subsequence in uniform topology. $\endgroup$
    – bellcircle
    Feb 9, 2017 at 3:12
  • $\begingroup$ @bellcircle the sequence is indeed not equicontinuous. Maybe $X$ is not compact? $\endgroup$ Feb 9, 2017 at 3:18
  • 1
    $\begingroup$ Yeah. I guess the problem has an error. $\endgroup$
    – bellcircle
    Feb 9, 2017 at 3:21
  • $\begingroup$ @bellcircle If there were a convergent subsequence it would also be one for the pointwise topology, which is weaker. And we already ruled that out. $\endgroup$ Feb 9, 2017 at 3:23
  • $\begingroup$ I also jumped right to Ascoli. But as Henno points out, it is easy to see that if the space is not compact in the pointwise convergence topology then it can't be compact in the finer one. $\endgroup$ Feb 9, 2017 at 3:43

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.

  • $\begingroup$ Ascoli says that any equicontinuous sequence has a convergent subsequence. Not every sequence has a convergent subsequence? $\endgroup$ Feb 9, 2017 at 3:20
  • $\begingroup$ The sup-metric is stronger, so if we 'd have compactness there, then also for the pointwise topology, quod non. $\endgroup$ Feb 9, 2017 at 3:28
  • $\begingroup$ you are right of course. $\endgroup$ Feb 9, 2017 at 3:37
  • $\begingroup$ I guess the point here then is to show that the topology of convergence on compact sets is the same as the sup norm topology. It's not hard, but it's not trivial either. $\endgroup$ Feb 9, 2017 at 3:44

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 $(I^I, \cal{T}_u)$.


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