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Building off a previous question, I'm trying to prove some properties about a certain function, but I may be flubbing the whole thing.

Suppose I have a well-ordered set $(B,\leq)$ where there is a unique $b\in B$ such that $\{a\in B\mid a\lt b\}$ is uncountable, but $\{a\in B\mid a\lt c\}$ is countable for all $c\neq b$.

With this set $(B,\leq)$ in mind, I can set up a base with the intervals of form $\{a\in B\mid a\lt x\}$, $\{a\in B\mid a\gt y\}$, and $\{a\in B\mid x\lt a\lt y\}$ for various $x,y\in B$. This covers $B$ and the intersection of two intervals is another interval (when the intersection is nonempty) so this family is indeed a base. I'll call it $\mathcal{B}$, and take the associated topology on $B$.

Now let $\phi\colon B\to \{0,1\}$ be defined as $\phi(b)=1$ and $\phi(a)=0$ for any $a\neq b$. Let $\{0,1\}$ have the discrete topology. So I believe $\phi$ is not continuous since $\phi^{-1}(\{1\})=\{b\}$ is not open in $B$ as $\{b\}$, as a singleton, cannot possibly have one of the three forms of intervals in $\mathcal{B}$. Am I correct in thinking this?

I want to finally show that for any convergent sequence $\{a_i\}$ in $B$, then $\lim_{i\to\infty}\phi(a_i)=\phi(\lim_{i\to\infty}a_i)$. How can one do this?

Let $a_i\to a$. My thinking is I just need to show any nbhd of $\phi(a)$ contains all but finitely many $\phi(a_i)$. So the only nbhds of $\phi(a)$ are $\{\phi(a)\}$ and $\{0,1\}$. Clearly $\{0,1\}$ contains all $\phi(a_i)$. If $\phi(a)=0$, then all but finitely many $a_i$ map to $\phi(a)$, since only $\phi(b)=1$. I think there might be more that needs to be said about this, my worry is that maybe $b$ shows up in the sequence infinitely many times?. And if $\phi(a)=1$, I'm not sure what to do. I'd appreciate an explanation on how this property holds. Many thanks.

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You might be interested to read my answer here, since your $B$ is exactly $\omega_1+1$. –  Asaf Karagila Sep 6 '11 at 22:07
    
I think you can generalize this , i.e., examples of how sequential continuity does not imply continuity for spaces that are not sequential, like any space with the cocountable topology. –  gary Sep 7 '11 at 5:57

4 Answers 4

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Assume that $a_i$ converges to $b$. I will prove that $a_i$ is constant $b$ from some point. We assume that there are infinitely many $a_i\ne b$ and seek a contradiction.

By definition of limit, every neighborhood of $b$ must contain the entire sequence from some point. In particular this is true for the neighborhood $(a,b]$ for each $a<b$ (upwardly "closed" intervals are allowed because $b$ has a unique successor unless it's the maximal element). Since we're assuming that the sequence does not end in infinite $b$s, this means that for every $a<b$ there is an $i$ such that $a<a_i<b$.

But this means that $$\bigcup_{i, a_i<b} \{ x \mid x<a_i \} \supseteq \{ x \mid x< b \}$$ The left-hand side is a countable union of countable sets, and must therefore be countable itself. But the right-hand-side is uncountable (as an assumed property of $b$), which is a contradiction.

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Thanks Henning Makholm, this proof iss quite useful. Do you have a corresponding argument for when the $a_i$ converge to something other than $b$? I'm going to attempt to work it out now, but would find it helpful so see one if you have one. –  Gotye Sep 6 '11 at 22:33
    
The other case is easy: If the $a_i$s converge to something different from $b$, there is a neighborhood of the limit that doesn't contain $b$, and therefore $\phi(a_i)$ is identically 0 from some point. –  Henning Makholm Sep 6 '11 at 22:35
    
Thanks, you are all too brilliant. I mulled this over last evening and this morning, get stuck, post question, and bam!, 3 nice answers in a matter of minutes. –  Gotye Sep 6 '11 at 22:43
    
Being familiar with ordinal numbers does help. :-) –  Henning Makholm Sep 6 '11 at 22:53

You are correct in thinking that your function $\phi$ is not continuous at $b$, for exactly the reason that you gave. For the main part of your question you have to distinguish two main cases: $a=b$, and $a<b$.

Exercise 1: Suppose that $a=b$. Show that there is some $n_0 \in \mathbb{N}$ such that $a_n = b$ for all $n \ge n_0$.

Exercise 2: Suppose that $a<b$. Show that there is some $n_0 \in \mathbb{N}$ such that $a_n \le a$ for all $n \ge n_0$.

Conclude that in both cases $\phi(\lim a_n)= \lim(\phi(a_n))$.

Your space, by the way, is essentially the space of ordinal numbers less than or equal to $\omega_1$, the first uncountable ordinal, with your $b$ corresponding to $\omega_1$; the topology is called the order topology.

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A minor typo in the last paragraph, I hope you did not mind. :-) –  Asaf Karagila Sep 6 '11 at 22:22
    
@Asaf: What was it? Did I leave off the subscript on one of the $\omega_1$’s? (I certainly don’t mind!) –  Brian M. Scott Sep 6 '11 at 22:28
    
Brian: "Your" was "You". You can click on edited xx time ago above Asaf's name and see what he did. –  t.b. Sep 6 '11 at 22:31
    
Thank you Brian M. Scott, I am going to attempt to figure these out. I hope you don't mind if I come back later if I get stuck. –  Gotye Sep 6 '11 at 22:34
    
@Theo: Thanks. I’m still learning my way around all the fancy machinery here. –  Brian M. Scott Sep 6 '11 at 22:35

Since your previous question revealed that $B=\omega_1+1$ with the usual $<$ of ordinals, and $b=\omega_1$ - I will use these notations instead as they are more common and familiar.

Some useful facts about ordinals:

  1. Every decreasing sequence of ordinals is finite.
  2. Every countable subset of $\omega_1$ is bounded.

This implies that the order topology on $\omega_1+1$ is compact, and it is not even first countable. That is, there is a point whose neighborhood basis is not countable - $\omega_1$.

The compactness follows from the argument that a covering by intervals you can write one of the intervals which covers $\omega_1$, then a lower interval meeting it, and continue. The lower point of each interval form a decreasing sequence of ordinals, which is finite. Therefore there is a finite sub-cover.

The limit of every sequence of countable ordinals is a countable ordinal. Furthermore, since every decreasing sequence of ordinals is finite, an infinite set of ordinals will have a strictly increase subset.

Therefore every increasing (countable) sequence of ordinals has a limit in $\omega_1+1$ (well, all the countable ordinals are in this space...). Therefore every sequence has a convergent subsequence.

Note that by the same argument if $\omega_1$ (which is not the limit of any non-constant sequence) is a member of a convergent sequence then it appears finitely many times.

Lastly, indeed the function that you have defined is not continuous, since as you said $\{\omega_1\}$ is not an isolated point, i.e. not an open set, this is because open sets are union of open intervals, and $\omega_1$ is a limit ordinal so every open interval would contain smaller ordinals (and in this case, uncountably many of them).

This space is a wonderful example for a compact space and not even first countable, as I remark in my answer here it can be made into an even more useful counterexample of notion of compactness, and countability axioms.

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Thank you Asaf Karagila, I wasn't expecting so much set theory to come up, but I will wade through this. –  Gotye Sep 6 '11 at 22:44
    
@Gotye: General topology is deeply connected with set theory. –  Asaf Karagila Sep 6 '11 at 22:55

Another example of how sequential continuity does not imply continuity:

Take the identity map from (R, cocountable) to (R, Std.) , where W is open in the Cocountable topology iff (def.) it has a countable complement , or if W is empty.

You can show that any sequence that converges in the cocountable topology is eventually-constant, so sequential continuity is trivial. But the identity map is not continuous; take any finite interval $(a,b)$ --i.e., $-\infty<a<b< \infty$ ; then $f^{-1}(a,b)=(a,b)$, but $\mathbb R- (a,b)=(-\infty,a]\cup[b,\infty)$ , which is not countable.

I think we can find examples for all non-sequential spaces.

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