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I am trying to show that $C[0,1],$ the space of all real - valued continuous functions with the sup metric is not sequentially compact with the sup metric by showing that the sequence $f_n = x^n$ has no convergent subsequence. The sup metric $\|\cdot\|$ is defined as

$$\|f - g \| = \sup_{x \in [0,1]} |f(x) - g(x)|$$

where $|\cdot|$ is the ordinary Euclidean metric. Now I know that $f_n \rightarrow f$ pointwise, where

$$f = \begin{cases} 0, & 0 \leq x < 1 \\ 1, & x = 1.\end{cases}$$

However $f \notin C[0,1]$ so this means by theorem 7.12 of Baby Rudin that $f_n$ cannot converge to $f$ uniformly. However how does this tell me that no subsequence of $f_n$ can converge to something in $C[0,1]$?

Thanks.

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    I think you can use the same argument for any subsequence: $f_{n_k}$ converges to $f$ pointwise.2012-06-17
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    @MartinSleziak If I know that any subsequence $f_{n_k}$ converges to $f$ pointwise, how does this mean that no subsequence can converge to something in $C[0,1]$?2012-06-17
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    If some sequence converges to something, then every subsequence converges to the same limit. If it would converge to $\hat{f}$ instead where $\|f-\hat{f}\|_\infty > 0$, then the whole sequence would not converge at all.2012-06-17
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    Maybe it helps to think in terms of single points to start with: Pick any $x\in [0,1].$ Pick *any* subsequence $f_{n_k}.$ The sequence $f_{n_k}(x)$ will converge to $f(x)$ since a subsequence of a convergent sequence converges to the same limit. So $f_{n_k}$ converges pointwise to $f$ which is not in $C[0,1]$. If a sequence of functions converges uniformly, it does so the the pointwise limit, so this subsequence can't converge in the sup metric to anything in $C[0,1].$2012-06-17
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    @BenjaminLim If $f_{n_k}\to g\in C[0,1]$ as $k\to\infty$, then $f_{n_k}\to g$ pointwise and hence $g=f$ on $[0,1]$. However, $f\not\in C[0,1]$, i.e., $f$ is not continuous, and thus we have a contradiction because $g$ is continuous by assumption.2012-06-17
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    @AmiteshDatta This thing about whether the function it converges to being in the space or not is getting me tied up in knots...2012-06-17
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    @BenjaminLim We can wait until your topology class to learn how to untie [knots](http://en.wikipedia.org/wiki/Knot_theory). =P2012-06-17
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    @BenjaminLim I think the following is an easier way to think about this. If $f_n\to f$ *in the metric space* $C[0,1]$, then $f_n\to f$ uniformly and conversely. In other words, convergence in $C[0,1]$ is the *same thing as uniform convergence*. However, irrespective of the existence of $C[0,1]$, we know that uniform convergence implies pointwise convergence. Therefore, if $f_n$ (or any subsequence of $f_n$) converges to $g$ *in $C[0,1]$*, then $f_n\to g$ pointwise ...2012-06-17
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    @BenjaminLim ... However, $f_n$ can converge to *at most one function* pointwise (this is the key point, I think, and depends on the uniqueness of limits in Hausdorff spaces because we are working pointwise). Therefore, $g=f$ because we know that $f_n\to f$ pointwise (this sentence and the last have nothing to do with the space $C[0,1]$). Of course, this is a contradiction because $f\not\in C[0,1]$ so it does not make sense to speak of whether $f_n\to f$ *in $C[0,1]$* (one requirement for convergence is that the limit is an element of the space!). Does that clarify the situation?2012-06-17
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    @AmiteshDatta I think I get it, when you say "Therefore $g = f$ because we know that $f_n \rightarrow f$ pointwise (this sentence and the last... $C[0,1]$) don't we somehow need to "enlarge" our space and say just consider the space of all functions on $[0,1]$? Namely because to talk about limits being unique,etc don't we need to live in one big *ambient space*?2012-06-17
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    @AmiteshDatta What if we go up to the space of all functions on $[0,1]$ that contains $C[0,1]$ with the sup metric? I think we then have a big ambient space to work in, and because we are working with convergent sequences, we don't need to worry about the sup metric blowing up or not.2012-06-17
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    @BenjaminLim You can also consider the space $B[0,1]$ of all real bounded functions on $[0,1]$ along with the sup metric and observe that $f \in B[0,1]$.2012-06-17
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    @BenjaminLim Yes, in a sense, we are viewing $f$ as an element of the (metric) space of all (bounded) functions of $[0,1]$. However, as I said, this part of the argument has nothing to do with $C[0,1]$ (or, indeed, any metric space); you can think of it as general nonsense with functions. Of course, arguments of this nature occur throughout mathematics, e.g., if you study Euclidean space and argue that compact subspaces are closed, then you are basically appealing to a general lemma in point-set topology regarding compact subspaces of Hausdorff spaces ...2012-06-17
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    @BenjaminLim ... In the argument answering your question, we are basically going up to a bigger space and applying arguments of a general kind there. I should add the technical point that the space of all functions on $[0,1]$ is **not** a metric space; as Sayantan argues, one needs to instead consider the space $B[0,1]$ of all real **bounded** functions on $[0,1]$ equipped with the supremum metric. The reason is that if $f$ is an unbounded function on $[0,1]$, then its distance to the zero function on $[0,1]$ is not well-defined.2012-06-17
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    @AmiteshDatta Yes that is correct, however if I want to argue that my $f_n$ converges to the discontinuous function in that space pointwise, don't I need to equip that space with the usual euclidean metric and not the sup metric?2012-06-17
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    @BenjaminLim However, in this argument, we are not using our understanding of the *metric space structure* of $B[0,1]$; rather, we are using its structure *purely as a set* because we simply wish to view $f$ as an element of a larger space. We don't need the metric space structure of $B[0,1]$ because we are only discussing pointwise convergence in this larger space which has nothing to do with the metric structure of $B[0,1]$ (except that convergence *in $B[0,1]$* implies pointwise convergence).2012-06-17
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    @AmiteshDatta In that case, how can I look at $C[0,1]$ as a subspace of $B[0,1]$ as they have different metrics? The one on $C[0,1]$, the sup metric is not simply the restriction of the euclidean metric to this space.2012-06-17
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    @AmiteshDatta We want to say limits are unique and such in $B[0,1]$, if I understand correctly don't we *need* to talk about the metric space structure then?2012-06-17
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    Why don't you just compute (or estimate) the maximum of $x^n - x^m$ on $[0,1]$ if $n \neq m$ and show that it is larger than $1/2$?2012-06-17
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    @BenjaminLim Yes, you would need to equip $\mathbb{R}$ with the structure of a metric space. In fact, you can speak of pointwise convergence of a sequence $f_n$ of functions on a topological space with values in *any metric space* (in this case $\mathbb{R}$ equipped with the standard metric). However, we are not explicitly referring to a metric on the *space of functions* when we discuss pointwise convergence; we only refer to a metric on the *range of the space of functions* and the range is $\mathbb{R}$ in this case.2012-06-17
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    @BenjaminLim We use the uniqueness of limits of sequences of **elements** in $\mathbb{R}$ to prove the uniqueness of **pointwise** limits of sequences of functions. Indeed, if $f_n\to f$ pointwise, then this means that $f_n(x)\to f(x)$ **for each** $x\in [0,1]$ as sequences of **elements** of $\mathbb{R}$ and the uniqueness of $f(x)$ follows from the fact that $\mathbb{R}$ is a Hausdorff space (note that we don't need any metric space different from $\mathbb{R}$ here). Since $x\in [0,1]$ was arbitrary, we conclude uniqueness of pointwise limits of sequences of **real-valued** functions.2012-06-17
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    @AmiteshDatta Please look at my answer that I have posted after discussion with you.2012-06-17
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    @t.b. I tried doing that to show that the sequence was not uniformly cauchy but it did not work out...2012-06-17
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    Then try harder :) If $n \gt m$ then $x^m \geq x^n$, so $x^m - x^n \geq 0$. The minimum is attained at $0,1$ and the maximum can be found by setting the derivative equal to zero: $mx^{m-1} - n x^{n-1} = 0$ if and only if $x^{m-1}(\frac{m}{n} - x^{n-m}) = 0$, plug in, done. It might help to remember [Bernoulli's inequality](http://en.wikipedia.org/wiki/Bernoulli's_inequality). This is as simple as it gets.2012-06-17
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    @t.b. $x-x^2=x(1-x)$ attains maximum at $x=1/2$ an the value is $1/4$, so it is not larger then $1/2$. Did I miss something there?2012-06-17
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    @t.b. That's exactly what I did. We want to show that there exists $\epsilon > 0$ such that for all $N$, there exist $m,n \geq N$ such that $||x^m - x^n ||<\epsilon$. Assume wlog that $n > m$, we then have by taking the derivative and stuff that the maxium value of the difference is achieved at the point $x$ that you calculated, and we get that $||x^m - x^n || = (m/n)^{m/(n-m)}\left(1 - \left(\frac{m}{n}\right)^{n-m}\right)$. However, how can I just choose an $\epsilon $ such that this expression is greater than $\epsilon$ for $m,n $ sufficiently large? I tried to fix $n$ and try taking $m$2012-06-17
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    sufficiently large but it did not work out.2012-06-17
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    @AmiteshDatta I think that by a corollary of the Arzelà - Ascoli theorem or some related result, because the family $\{f_n\}$ is not equicontinuous at $x = 1$, we have that $\{f_n\}$ is not a compact subset of $C[0,1]$. In particular I think this means that no subsequence of $f_n$ can every converge to anything in the space.2012-06-17

4 Answers 4

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After all of this discussion, here is my understanding of how the problem concludes: We already know that $f_n \rightarrow f$ pointwise, where $f$ is the discontinuous function that I defined on $C[0,1]$. Now go up to a bigger metric space like $B[0,1]$ that contains $C[0,1]$ and equip it with the sup metric. In there we know that $f_n \rightarrow f$ pointwise. This means that for each $x \in [0,1]$, the ordinary sequence of real numbers $f_n(x) \rightarrow f(x)$ and since $\Bbb{R}$ with the Euclidean topology is Hausdorff it follows that the pointwise limit of $f(x)$ is unique. Viz, there is no other function on $B[0,1]$ that converges to $f$ pointwise.

Now I claim that there is no function $g \in C[0,1]$ for which $f_n$ converges to uniformly. If there were, then $f_n \rightarrow g$ pointwise on $C[0,1]$. Furthermore, a continuous real valued function on a compact set is bounded and so $g$ lives in $B[0,1]$. By what I said about limits being unique, this means that $g = f$. However this is a contradiction because $f \notin C[0,1]$.

It follows that there is no continuous function on $[0,1]$ that $f_n$ converges to. Q.E.D.

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    I have a couple of corrections/comments. Firstly, as I said before, you don't need to equip $B[0,1]$ with the supremum metric because the metric structure of $B[0,1]$ is irrelevant to the problem at hand. Indeed, we are studying pointwise convergence and we only need to know that $C[0,1]\subseteq B[0,1]$ *as sets* but not *as metric spaces*; indeed, I emphasize that we don't care about any metric space structure on $B[0,1]$ for the purposes of this problem. We only need $B[0,1]$ as a convenient set in which the function $f$ lives ...2012-06-17
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    ... Secondly, I think your sentence "... there is no other function on $B[0,1]$ that converges to $f$ pointwise ..." doesn't make sense; a more meaningful sentence might be "... if $g\in B[0,1]$ is such that $f_n\to g$ pointwise, then $f=g$ ...". I emphasize that there is nothing "special" about $B[0,1]$ in this regard; the only relevant property is that $f\in B[0,1]$ and it doesn't matter that "$B[0,1]$" even means as long as it is a set, $C[0,1]\subseteq B[0,1]$, and $f\in B[0,1]$ ...2012-06-17
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    ... In other words, my point is that you're making the problem more complicated by introducing $B[0,1]$. A simpler argument is to note the uniqueness of pointwise limits; this involves the metric space structure of $\mathbb{R}$ (notice $B[0,1]$ is not relevant here). You can then finish the argument by noting that the pointwise limit of any subsequence of $\{f_n\}_{n\in\mathbb{N}}$ is equal to $f$ (notice $B[0,1]$ is not relevant here) ...2012-06-17
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    ... Of course, this leads to a contradiction because uniform convergence implies pointwise convergence and $f$ is not continuous (notice $B[0,1]$ is not relevant here). You can introduce $B[0,1]$ if you want but even if you do, you shouldn't need to care about any metric space structure on $B[0,1]$. However, the solution is much simpler if you forget about $B[0,1]$.2012-06-17
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    My "ultimate" point is that you can speak of pointwise convergence of *any sequence of functions* wherever they live provided that their (common) range is a metric space; this doesn't involve any metric space structure (especially any metric space structure on a $B[0,1]$ (even if we care what that means which we don't)) except the metric space structure on the (common) range of the sequence of functions.2012-06-17
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    @AmiteshDatta Dear Amitesh, I have thought about this problem and it actually is very simple. I am getting tied up in knots. Regards,2012-06-18
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    Dear Ben, that happens to everyone; good to hear that it's clear now. Regards,2012-06-19
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Since $f_n$ converges to $f$ pointwise, all subsequences $f_{n_k}$ also converge pointwise to $f$. So $f_{n_k}$ can not converge uniformly to continuous $g$, because uniform convergence implies pointwise convergence as well, and $f$ is not continuous.

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This is (more-or-less) a modification of the hint from t.b's comment.

It suffices to show that any subsequence of $(f_n)$ is not Cauchy. (Since every convergent sequence is a Cauchy sequence.)

To do this, notice that $$\lVert f_{2n}-f_n \rVert = \sup_{x\in[0,1]} (x^n - x^{2n}) = \sup_{x\in[0,1]} (x^n - (x^{n})^2) = \sup_{t\in[0,1]} (t-t^2)=\frac14.$$ (It is easy to find maximum of the quadratic function $f(t)=t-t^2=t(1-t)$.)

In addition, if we use the monotonicity of the sequence $(f_n)$, we see that $$k\ge 2n \qquad \Rightarrow \qquad \lVert f_{k}-f_n \rVert \ge \frac14.$$

Now, if we have any subsequence of $(f_n)$ then the above estimate shows that this subsequence is not Cauchy. For any given $k_0$ we can find $k'>k_0$ such that $n_{k'}>2n_{k_0}$ and $\lVert f_{n_{k'}}-f_{n_{k_0}} \rVert \ge \frac 14$.

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Regarding the motivation for your question (sequential compactness):

First, commenting on your specific question: a topological vector space over the real or complex numbers is never sequentially compact. Only bounded closed subset are possibly (sequentially) compact. Your attempt of proof is, therefore, only relevant if you restrict to bounded sequences (which applies to your example but should be made an explicit assumption).

As a general remark: it is known that compactness and sequential compactness are equivalent in metric spaces, hence in subsets of normed vector spaces (like $C^0$).

It is also known (though a bit more involved when it comes to proving it) that a normed (real or complex) vector space has the so called 'Heine Borel' property (which says that a subset is compact if and only if it is closed and bounded) if and only if it is of finite dimension.

Edit: the following sentence is obviouly not correct (see the comment of t.b. -- thanks for pointing this out), but is a nice illustration of a premature und uncautios conclusion, so I don't delete it ;-): Since $C^0$ is not finite dimenional, it follows that bounded closed sets of $C^0$ are (sequentially) compact only if they are contained in a finite dimensional subspace.

The following remains true, though: sets like the closed unit ball (and therefore any set containing an open ball) are not (sequentially) compact in $C^0$ and, in general, bounded sequences need not have a converquent subsequence, and this applies to every infinite dimensional normed vector space.

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    The set $\{0\} \cup \{\frac{1}{n}x^n\,:\,n \geq 1\}$ *is* compact in $C[0,1]$, hence bounded and closed but it is not contained in a finite-dimensional subspace...2012-06-17
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    @t.b.: thank you for pointing this out. I edited my posting.2012-06-17