Challenge Aren't you tired of counterexamples already?

[micromass]

I have a question here: Shall the collection of pairwise disjoint open sets itself be countable or the index set of it?

What is the difference?

 

[fresh_42]

What is the difference?

$O_x ∪ O_y$ are countably many open sets whereas each single one must not contain countably many elements. I suppose the index set, here $\{x,y\}$ is supposed to be countable.

 

[micromass]

$O_x ∪ O_y$ are countably many open sets

What you wrote down is one open set. I don't get it.

 

[fresh_42]

What you wrote down is one open set. I don't get it.

$]0,1[$ is a union of countably many open sets (1), whereas $]0,1[$ has uncountably many elements so its union isn't countable, too. So does the "countable" in 8. refer to the open sets itself or the cardinality of the set which indexes the union which I assume?
(There are certainly topologies in which open sets can have countably many elements.)

 

[micromass]

$]0,1[$ is a union of countably many open sets (1), whereas $]0,1[$ has uncountably many elements so its union isn't countable, too. So does the "countable" in 8. refer to the open sets itself or the cardinality of the set which indexes the union which I assume?
(There are certainly topologies in which open sets can have countably many elements.)

It's the cardinality of the index set.

 

[Samy_A]

A little off topic, but doesn't warrant a new thread, I think.

Does this work?
$A \subset \mathbb R²$ is separable.
That means that there exist a countable $B \subset A$ such that $\bar B =A$.
$B$ being countable means that $\mathring B = \varnothing$, so that $\partial B = \bar B \setminus \mathring B = A$.

Why is $A$ separable? I'll accept that $\mathbb{R}^2$ is separable.

The reasoning was: $\mathbb R²$ is second-countable, and I thought that implied that every subspace also is second-countable. And second-countable implies separable.

OK, that's good enough for me.

While I think the reasoning is correct, I nevertheless tried to prove directly that A is separable.
What I get are really variations on the proof that second-countable implies separable.
One example, very similar to the proofs I tried to concoct, can be found here:

Given: A second-countable space $X$, with countable basis $\{B_n\}$.
To prove: There exists a countable dense subset of $X$
Proof: We can assume without loss of generality that all the $\{B_n\}$ are nonempty, because the empty ones can be discarded. Now, for each $B_n$, pick any element $x_n \in B_n$. Let $D$ be the set of these $x_n$. $D$ is clearly countable (because the indexing set for its elements is countable). We claim that $D$ is dense in $X$.

To see this, let $U$ be any nonempty open subset of $X$ Then, $U$ contains some $B_n$, and hence, $x_n \in U$. But by construction, $x_n \in D$, so $D$ intersects $U$, proving that $D$ is dense.

My question: I think the axiom of choice is needed to construct $D$. Is this correct?

 

[micromass]

My question: I think the axiom of choice is needed to construct $D$. Is this correct?

Yes, in the proof you gave, this appears to be the case. But maybe there is a way to modify the proof so that the axiom of choice is not needed.

 

[fresh_42]

Yes, in the proof you gave, this appears to be the case. But maybe there is a way to modify the proof so that the axiom of choice is not needed.

The proof I've found in my textbook also uses The Axiom. I think otherwise you can't achieve something dense and countable under such a general condition.

 

[Samy_A]

Yes, in the proof you gave, this appears to be the case. But maybe there is a way to modify the proof so that the axiom of choice is not needed.

The proof I've found in my textbook also uses The Axiom. I think otherwise you can't get rid of the uncountability.

Well, the proof I posted "only" uses the axiom of countable choice.
I thought I had found a proof without that, but no luck. Somewhere in my proof I used that a countable union of finite sets is countable. And that apparently also needs the axiom of countable choice.

 

[fresh_42]

Well, the proof I posted "only" uses the axiom of countable choice.
I thought I had found a proof without that, but no luck. Somewhere in my proof I used that a countable union of finite sets is countable. And that apparently also needs the axiom of countable choice.

Have you tried to convert it into a proof on induction? (Sorry, I may have as well, but I need a short break.)

 

[Samy_A]

Have you tried to convert it into a proof on induction? (Sorry, I may have as well, but I need a short break.)

No, thanks for the idea.
Not that it is a big deal. I was just mildly surprised that such a well known fact of topology, second-countable⇒separable, depends on the axiom of countable choice, and that I never noticed that. Or maybe I forgot: it has been quite a few years since I learned topology.

 

[mfb]

5[/color] looks trivial:
=> If a function is continuous, it is equal to a continuous function (itself) almost everywhere.
<= Counterexample: f(0)=1, f(x)=0 elsewhere is equal to g(x)=0 almost everywhere, but not continuous.

1[/color]:
<= Counterexample: Consider the sequence 1, 1/4, 1/4, 1/16, 1/16, 1/64, ... The ratio test will be inconclusive because the ratio between adjacent elements does not converge. The root test, however, will show that $\sqrt[n]a_n \leq \frac 1 2$.
=> If the ratio test is conclusive, then the root test will be so as well. WLOG assume all a_n are positive because the ratio test gives the stronger absolute convergence, this just saves writing | | everywhere. Let the limit of the ratios be $Q=1-2\epsilon$. Then there is an N such that $\frac{a_{n+1}}{a_n} < 1-\epsilon\quad \forall n>N$. Restrict the analysis to n>N. Then $a_n \leq a_N (1-\epsilon)^{n-N}$. Take the nth root: $\sqrt[n]{a_n} \leq \sqrt[n]{a_N (1-\epsilon)^N} \sqrt[n]{(1-\epsilon)^n} = \sqrt[n]{a_N (1-\epsilon)^N}(1-\epsilon)$. The remaining root converges to 1, therefore the expression will converge to $1-\epsilon$ and the root test works.Edit: and a double counterexample for #9[/color]:

Take an arbitrary non measurable function f:R->R, but set f(x)=0 for 0<x<1. It is then differentiable of arbitrary order somewhere (e. g. at x=0.5).

g(x)=0 for x rational, g(x)=1 for x irrational is measurable, but nowhere differentiable.

 

[fresh_42]

I really regret that I haven't worked out my copy of Hewitt, Stromberg ...

 

[mfb]

micromass, did you see my edit in the previous post?

And did someone find the "both directions are right" already?

 

[Samy_A]

And did someone find the "both directions are right" already?

That's 3.
Joint effort by @fresh_42 and me.

 

[Samy_A]

A compact topological space $X$ is separable if and only if each collection of pairswise disjoint open sets is countable.

The counterexample must be some weird topology.
Assume $X$ is compact, and $(U_i)_i$ is an uncountable family of pairswise disjoint open sets. If $U=\cup U_i$, then no way a open cover of $U$ can have a finite subcover. So there must "something " weird going on in the complement of $U$, making $X$ compact.

So, take $X= \mathbb R$. Define the following topology on $X$.
Sets that don't include the point 0 are open;
for sets that include the point 0, we take the cofinite topology, meaning that such a set is open if and only if its complement is finite.

That is a legitimate topology. $X$ is open as the complement of $X$ is finite. $\varnothing$ is open as it doesn't contain 0.
A finite intersection of open sets is open, as finite intersections of cofinite sets are cofinite.
Since a set larger than a cofinite set is cofinite, the union of open sets is open.
If the union or finite intersection doesn't include an open set that includes 0, then it is open, as outside 0 we have the discrete topology.

X is compact, since any open cover must include a neighborhood of 0, and the complement of that one is finite, making the existence of a finite subcover trivial.
X is not separable, since no countable set can intersect all the points outside 0.
And of course $\mathbb R \setminus \{0\}$ is an uncountable collection of pairswise disjoint open sets, namely the singletons.

The proof of the other direction is elementary.
If X is separable, it contains a countable dense set. That set cannot intersect every element of an uncountable collection of pairswise disjoint open sets. Hence a collection of pairswise disjoint open sets must be countable.

 

[micromass]

The counterexample must be some weird topology.
Assume $X$ is compact, and $(U_i)_i$ is an uncountable family of pairswise disjoint open sets. If $U=\cup U_i$, then no way a open cover of $U$ can have a finite subcover. So there must "something " weird going on in the complement of $U$, making $X$ compact.

So, take $X= \mathbb R$. Define the following topology on $X$.
Sets that don't include the point 0 are open;
for sets that include the point 0, we take the cofinite topology, meaning that such a set is open if and only if its complement is finite.

That is a legitimate topology. $X$ is open as the complement of $X$ is finite. $\varnothing$ is open as it doesn't contain 0.
A finite intersection of open sets is open, as finite intersections of cofinite sets are cofinite.
Since a set larger than a cofinite set is cofinite, the union of open sets is open.
If the union or finite intersection doesn't include an open set that includes 0, then it is open, as outside 0 we have the discrete topology.

X is compact, since any open cover must include a neighborhood of 0, and the complement of that one is finite, making the existence of a finite subcover trivial.
X is not separable, since no countable set can intersect all the points outside 0.
And of course $\mathbb R \setminus \{0\}$ is an uncountable collection of pairswise disjoint open sets, namely the singletons.

The proof of the other direction is elementary.
If X is separable, it contains a countable dense set. That set cannot intersect every element of an uncountable collection of pairswise disjoint open sets. Hence a collection of pairswise disjoint open sets must be countable.

Don't you want to find a topology where every disjoint collection of open sets is countable but that is not separable? Your example is called the Fort space, it is the Alexandroff compactification of a discrete space. https://en.wikipedia.org/wiki/Fort_space

 

[Samy_A]

Damn yes!

 

[Samy_A]

Amending the previous post, I define the topology as follows:
$X= \mathbb R$
$\mathbb R, \varnothing$ are open.
Sets that don't include 0 are open if their complement is countable.
Sets that include 0 are open if their complement is finite.

X is compact, since any open cover must include a neighborhood of 0, and the complement of that one is finite, making the existence of a finite subcover trivial.
X is not separable, since given a countable set $D$, $\mathbb R \setminus (D \cup \{0\})$ is open and doesn't intersect $D$.

If $U,V$ are non empty open sets, then their complement is countable. Hence $U \cap V$ also has a countable complement, and cannot be empty. Hence there are no uncountable families of pairswise disjoint open sets.
Not sure that qualifies, but worth a try.

EDIT: I think I could also take the Alexandroff compactification of the cocountable topology on $\mathbb R$. Essentially the same arguments hold (if they are valid, that is).

 

[micromass]

Yes, this one seems right!

 

[andrewkirk]

2. In any Banach space $X$ and given any series $\sum a_n$ in $X$, then the series converges absolutely (that is: $\sum \|a_n\|$ converges) if and only if the series converges unconditionally (that is: for any bijection $\pi:\mathbb{N}\rightarrow \mathbb{N}$ holds that $\sum_n a_{\pi(n)}$ converges to the same number).

I think I can prove the forward direction.

Let $a_n$ be a sequence whose series is absolutely convergent to $A$.
Let $\sigma:\mathbb N\to\mathbb N$ be a bijection. We claim that $\sum_{n=1}^\infty a_{\sigma(n)}=A$.

Attempted Proof:
Define the following :

$A_n=\sum_{k=1}^n a_n$
$S_n=\sum_{k=1}^n\|a_n\|$
$S=\lim_{n\to\infty}s_n$. We know this limit exists because the series absolutely converges.
$A'_n=\sum_{k=1}^n a_{\sigma(n)}$

Given $\epsilon>0$ choose $N\in\mathbb N$ such that for any $n\geq N$ both $\|S-S_n\|$ and $\|A-A_n\|$ are less than $\frac\epsilon9$.

Let $M=\max\{\sigma^{-1}(k)\ |\ k\leq N\}$.

We claim that $n\geq M\Rightarrow \|A'_n-A\|<\epsilon$, which is sufficient to prove the forward direction.

Let $n$ be any integer that is $\geq M$, and let $B$ be the set of integers in $\{1,...,n\}$ that are not in $\sigma^{-1}\left(\{1,...,N\}\right)$.

Here goes:

\begin{align*}
\|A'_n-A\|&=\|(A_N+\sum_{k\in B}a_{\sigma(k)})-A\|\\
&\leq \|A_N-A\|+\|\sum_{k\in B}a_{\sigma(k)}\|\\
&\leq \frac\epsilon9+\sum_{k\in B}\|a_{\sigma(k)}\|\\
&\leq \frac\epsilon9+\sum_{k>N}\|a_k\|\\
&= \frac\epsilon9+\|S-S_N\|\\
&\leq\frac\epsilon9+\frac\epsilon9<\epsilon
\end{align*}

That seemed quite easy (although it might be because there's a mistake in there). If so then I'm guessing that the reverse direction will be the harder one. I'll need to look at that later.

Edit: corrected typo in first line of aligned equations, pointed out by Samy (see post below)

 

[Samy_A]

Here goes:

\begin{align*}
\|A'_n-A\|&=\|(A_n+\sum_{k\in B}a_{\sigma(k)})-A\|\\
&\leq \|A_n-A\|+\|\sum_{k\in B}a_{\sigma(k)}\|\\
&\leq \frac\epsilon9+\sum_{k\in B}\|a_{\sigma(k)}\|\\
&\leq \frac\epsilon9+\sum_{k>N}\|a_k\|\\
&= \frac\epsilon9+\|S-S_N\|\\
&\leq\frac\epsilon9+\frac\epsilon9<\epsilon
\end{align*}.

Is there a typo in the first line? $A_n$ should be $A_N$, I think.

That seemed quite easy (although it might be because there's a mistake in there). If so then I'm guessing that the reverse direction will be the harder one. I'll need to look at that later.

This was the difficult part.

 

[andrewkirk]

This was the difficult part.

Ah, silly me, I didn't read the instructions in the OP closely enough. I was looking to prove the opposite direction, but all I needed was a counterexample to disprove that direction. I couldn't see that anybody had done that yet above (maybe I missed it). So:

Consider the Hilbert space $l^2$. Let $b_1,b_2,...$ be such that $b_k$ is the sequence that is all zeroes except for a 1 in the $k$th place. It is easy to show that this is an orthonormal basis for the space. Then the sequence $a_k=\frac{b_k}k$ is not absolutely convergent, as
\begin{align}\sum_{k=1}^\infty \|a_k\|=
\sum_{k=1}^\infty \frac1k\end{align}
which does not converge.
But, given $\epsilon>0$ we can choose $N$ such that if $n\geq N\Rightarrow \sum_{k=n}^\infty\frac1{k^2}<\frac{\epsilon^2}9$. So, for $m,n\geq N$ we have (defining $S_k\equiv\sum_{k=1}^n s_n$):
\begin{align}
\|S_m-S_n\|&=\left\|\sum_{j=n+1}^m a_j\right\|\\
&=\sqrt{\left\langle \sum_{j=n+1}^m a_j,\sum_{j=n+1}^m a_j\right\rangle}\\
&=\sqrt{\sum_{j,k=n+1}^m \langle a_j,a_k\rangle}\\
&=\sqrt{\sum_{j,k=n+1}^m \frac{\delta_{jk}}{jk}}\\
&=\sqrt{\sum_{k=n}^\infty \frac1{n^2}}<\frac\epsilon3<\epsilon\\
\end{align}
So the sequence $S_n$ is Cauchy and, since the space is complete, must converge, to some point $p$. Choose $R$ such that $n\geq R\Rightarrow \|S_n-p\|<\frac\epsilon9$.

Now let $\sigma:\mathbb N\to\mathbb N$ be any rearrangement and set

$M=\max\{\sigma^{-1}(k)\ |\ k\leq N\}$

$S'_k=\sum_{j=1}^n a_{\sigma(j)}$

$H\equiv\max(R,M,N)$

$n\geq H$

$B(n)=\sigma(\{1,...,n\})-\{1,...,H\}$

Then
\begin{align*}
\|S'_n-p\|&\leq \|S'_n-S_n\|+\|S_n-p\|\\
&< \|S'_n-S_H\|+\|S_H-S_n\|+\frac\epsilon9\\
&= \left\|\sum_{k\in B(n)}a_k\right\|+\frac\epsilon9+\frac\epsilon9\\
&= \sqrt{\left\langle\sum_{k\in B(n)}a_k,\sum_{k\in B(n)}a_k\right\rangle}+\frac{2\epsilon}9\\
&= \sqrt{\sum_{\substack{j,k\in B(n)}}\left\langle a_j,a_k\right\rangle}+\frac{2\epsilon}9\\
&= \sqrt{\sum_{\substack{j,k\in B(n)}}\frac{\delta_{jk}}{jk}}+\frac{2\epsilon}9\\
&= \sqrt{\sum_{\substack{k\in B(n)}}\frac1{k^2}}+\frac{2\epsilon}9\\
&\leq \sqrt{\sum_{\substack{k\in B(n)}}\frac1{k^2}}+\frac{2\epsilon}9\\
&\leq \sqrt{\sum_{\substack{k=N+1}}^\infty\frac1{k^2}}+\frac{2\epsilon}9\\
&<\frac\epsilon3+\frac{2\epsilon}9=\frac{5\epsilon}9<\epsilon\\
\end{align*}

So the re-arranged series $S'_n$ converges to $p$ and hence the series is unconditionally convergent. Hence the sequence $a_n$ is a counter-example to the reverse direction.

 

[micromass]

Hint for (7): every pointswise limit of continuous functions has only countably many discontinuity points. Use Baire's theorem.

 

[ProfuselyQuarky]

Hint for (7): every pointswise limit of continuous functions has only countably many discontinuity points. Use Baire's theorem.

Can you post the counterexample since it seems like no one is going to do this one? I've been waiting to see it.

 

[micromass]

I will give credit to Samy_A for question $7$ since he came up with an answer to the question before I revised it. His answer still answers one direction too.

So, here's the result for $7$. On one side, continuous functions are Borel measurable. Limits of Borel measurable functions are Borel measurable. This the limit of a continuous limit is definitely Borel measurable.

On the other hand, not all Borel measurable functions are the limit of a sequence of continuous functions. One counterexample here is the indicator function of $\mathbb{Q}$. The idea is that this function has "many" many discontinuity points, indeed: every point of $\mathbb{R}$ is a discontinuity point. I will prove now however that if $f = \lim_n f_n$ is a limit of continuous functions then $f$ has a discontinuity set that is of first category.

Some terminology:
A closed set is called nowhere dense if its interior is empty. A set is called of first category if it is contained in the countable union of nowhere dense sets. A set is called of second category if it is not of first category.
For any function $f:U\rightarrow \mathbb{R}$ define $\text{osc}_g(x_0) = \lim_{\delta\rightarrow 0} \text{sup}_{|x-x_0|<\delta} |g(x) - g(x_0)|$. This is called the oscillation of $g$ at $x_0$. Note that $g$ is continuous at $x_0$ if and only if $\text{osc}_g(x_0) = 0$.

Lemma: If $U\subseteq \mathbb{R}$ is open and $\varepsilon>0$, then
$$\overline{\{x\in U~\vert~\text{osc}_g(x)>2\varepsilon\}}\subseteq \{x\in U~\vert~\text{osc}_g(x)\geq \varepsilon/2\}$$

Proof: We need to see that the limit points of the set on the left are in the set on the right. Thus suppose that $\text{osc}_g(x_n)\geq 2\varepsilon$ for all $n$ and that $x_n\rightarrow x_0$. For each $n$, choose $x_{n,m}$ such that $\lim_m x_{n,m} = x_n$ and let $|g(x_{n,m}) - g(x_n)|\geq \varepsilon$ for all $m$. Because convergence of $x_{n,m}$ to $x_n$, we may choose, for each $n$, an integer $m_n$ such that $|x_{n,m_n} - x_n| < |x_0-x_n|$ and then $\lim_n x_{n,m_n} = x_0$ by the triangle inequality. From $|g(x_{n,m_n}) - g(x_n)|\geq \varepsilon$, the triangle inequality forces
$$|g(x_{n,m_n}) - g(x_0)|\geq \varepsilon/2~\text{or}~|g(x_n) - g(x_0)|\geq \varepsilon/2.$$
Defining $y_n$ to be $x_{n,m_n}$ or $x_n$ according to the first or second inequality, we have $y_n\rightarrow x_0$ and $|g(y_n) - g(x_0)|>\varepsilon/2$. This proves the lemma.

In the following, we will use the Baire category theorem which states that $\mathbb{R}$ is of second category. Furthermore, all nonempty open subsets of $\mathbb{R}$ are of second category.

Theorem: If $f_n:\mathbb{R}\rightarrow \mathbb{R}$ are continuous functions converging pointswise to a function $f:\mathbb{R}\rightarrow \mathbb{R}$, then the set of discontinuous is of the first category.

Proof: In view of the lemma and the fact that $\mathbb{R}$ is not of the first category, it is enough to prove for each $\varepsilon>0$, that $\{x\in \mathbb{R}~\vert~\text{osc}_f(x)\geq \varepsilon\}$ does not contain a nonempty open subset of $\mathbb{R}$. Assuming the contrary, suppose that it contains the nonempty open set $V$. Define
$$A_{mn} = \{x\in V~\vert~|f_m(x) - f_n(x)|\leq \varepsilon/4\}.$$
This is closed in $V$. Then $A_m = \bigcap_{n\geq m} A_{mn}$ is closed in $V$. If $x\in V$, then the fact that $\{f_n(x)\}$ is Cauchy implies that there is some $m$ such that $x\in A_{mn}$ for $n\geq m$. Hence $\bigcup_m A_m = V$. The Baire theorem implies that some $A_m$ has nonempty interior. Fix that $m$ and let $W$ be its nonempty interior. Since
$$A_m\subseteq \{x\in V~\vert~|f_m(x) - f(x)|\leq \varepsilon/4\},$$
every point of $W$ has $|f_m(x) - f(x)|\leq \varepsilon/4$ and $\text{osc}_f(x)\geq \varepsilon$. Let $x_0\in W$ and choose $x_n\rightarrow x_0$ wth $|f(x_n) - f(x_0)|\leq 3\varepsilon/4$. From $|f_m(x_n) - f(x_n)|\leq \varepsilon/4$ and $|f_m(x_0) - f(x_0)|\leq \varepsilon/4$, we obtain $|f_m(x_n) - f_m(x_0)|\geq \varepsilon/4$. Since $x_n$ converges to $x_0$, this inequality contradicts the continuity of $f_m$ at $x_0$. This proves the theorem.

Finally, the characteristic function has entire $\mathbb{R}$ as discontinuity set, but this is not of first category by the Baire category theorem.

For more information, see https://en.wikipedia.org/wiki/Baire_function and Knapp's "Basic real analysis" from which I adapted this proof.