A function is not continuous, but the image of convergent sequences converge 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.
 A: 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:


*

*Every decreasing sequence of ordinals is finite.

*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.
A: 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.
A: 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.
A: 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. 
