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Theorem

A function $f:X\rightarrow Y$ is continuous at $x_0\in X$ if and only if any net $x_\lambda$ converging to $x_0$ is such that $f(x_\lambda)$ converges to $f(x_0)$.

Clearly any sequence is a net so that by last theorem I argue that a function $f:X\rightarrow Y$ is continuous at $x_0\in X$ if and only if any sequence $x_n$ converging to $x_0$ is such that $f(x_n)$ converges to $f(x_0)$ anway my text says that this is true only for first contable space so that I argue that notwithstanding $x_n\rightarrow x$ imply that $f(x_n)\rightarrow f(x)$ it could be that there exist a net $x_\lambda$ from a dire cted set $\Lambda$ different to $\Bbb{N}$ such that $x_\lambda\rightarrow x_0$ but $f(x_\lambda)\nrightarrow f(x_0)$? So could someone explain better this point with a counterexample?

Antonio Maria Di Mauro
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1 Answers1

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The problem is that if a space isn’t first countable, its topology may not be completely determined by its convergent sequences. Here’s a relatively straightforward example.

Let $\tau$ be the order topology on $\omega_1+1$, and let

$$\tau'=\big\{\{\alpha\}:\alpha\in\omega_1\big\}\cup\big\{\{\omega_1\}\cup(\omega_1\setminus C):C\subseteq\omega_1\text{ and }|C|\le\omega\big\}\cup\{\varnothing\}\;;$$

$\tau'$ is a topology on $\omega_1$ finer than $\tau$. (In other words, we just isolate each $\alpha\in\omega_1$ and let the point $\omega_1$ keep its usual nbhds.) Let $X$ be $\langle\omega_1+1,\tau'\rangle$ and $Y$ be $\langle\omega_1+1,\tau\rangle$.

Note that the only convergent sequences in $X$ are the ones that are eventually constant, so any function from $X$ to $Y$ takes convergent sequences in $X$ to convergent sequences in $Y$; it will not be surprising, then, that some of those functions are not continuous. Now let

$$f:X\to Y:\alpha\mapsto\begin{cases} 0,&\text{if }\alpha<\omega_1\\ \omega_1,&\text{otherwise;} \end{cases}$$

then $Y\setminus\{0\}$ is open in $Y$, but its inverse image under $f$ is $\{\omega_1\}$, which is not open in $X$, so $f$ is not continuous.

Brian M. Scott
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  • It seems to me that $\tau'$ is the discrete topology because if ${{\alpha}:\alpha\in\omega_1}\subseteq\tau'$ then any subset of $\omega_1$ is open, where is my mistake? – Antonio Maria Di Mauro Jun 28 '20 at 20:52
  • @AntonioMariaDiMauro: You forgot the point $\omega_1$: its nbhds are cocountable sets containing it. The other points of $X$ are indeed isolated. – Brian M. Scott Jun 28 '20 at 20:54
  • Okay, I roughly realise. However I don't understand why the only convergent sequences in $X$ are the ones that are eventually constant: indeed if $\alpha_n$ is a sequence in $X$ that converges to $\alpha$ then if $\alpha<\omega_1$ there must exist $n_0\in\Bbb{N}$ such that $\alpha_n\in{\alpha}$ for any $n\ge n_0$ so that $\alpha$ it looks to me semiconstant but not always constat and then what happens if $\alpha_n$ converges to $\omega_1$? – Antonio Maria Di Mauro Jun 28 '20 at 21:11
  • @AntonioMariaDiMauro: Eventually constant means that there is an $n_0\in\omega$ such that $\alpha_n=\alpha_{n_0}$ for all $n\ge n_0$; in other words, the sequence is constant from some point on. And the only sequences that converge to $\omega_1$ are constant at $\omega_1$ from some point on. – Brian M. Scott Jun 28 '20 at 21:13
  • Okay, so your eventually constant is the same that my semiconstat. However is $\alpha_n$ converges to $\omega_1$ then for any countable (eventually infinite) subset $C$ of $\omega_1$ there exist $n_0\in\Bbb{N}$ such that $\alpha_n\in{\omega}\cup(\omega_1\setminus C)$ so how can I argue that $\alpha_n=\omega_1$ for any $n\ge n_0$? Fogive my confusion. – Antonio Maria Di Mauro Jun 28 '20 at 21:18
  • Perhaps I understood. – Antonio Maria Di Mauro Jun 28 '20 at 21:29
  • So if we consider the neighborhoods ${\omega_1}\cup\omega$ and ${\omega_1}\cup(\omega_1\setminus\omega)$ of $\omega_1$ and if $\alpha_n\neq\omega_1$ for any $n\in\Bbb{N}$ but $\alpha_n$ converges to $\omega_1$ then there exist $n_0,n'_0\in\Bbb{N}$ such that $\alpha_n\in\omega$ for any $n\ge n_0$ and $\alpha_n\in\omega_1\setminus\omega$ for any $n\ge n'_0$ and so $\alpha_n\in(\omega_1\setminus\omega)\cap\omega$ for any $n\ge\max{n_0,n'_0}$ and this clearly is impossible so that there exist $n_0\in\Bbb{N}$ such that $\alpha_n=\omega_1$ for any $n\ge n_0$, is my a good explanation? – Antonio Maria Di Mauro Jun 28 '20 at 21:29
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    @AntonioMariaDiMauro: If $\langle\alpha_n:n\in\omega\rangle$ converges to $\omega_1$, let $M={n\in\omega:\alpha_n\ne\omega_1}$, and let $C={\alpha_n:n\in M}$. Then $C$ is countable, so ${\omega_1}\cup(\omega\setminus C)$ is an open nbhd of $\omega_1$, so there must be an $n_0\in\omega$ such that $\alpha_n=\omega_1}$ whenever $n_0\le n\in\omega$. Thus, the sequence must be eventually constant at $\omega_1$. – Brian M. Scott Jun 28 '20 at 21:38
  • Okay, now it is all clear. Thanks too much for your assistance! – Antonio Maria Di Mauro Jun 28 '20 at 21:41
  • @AntonioMariaDiMauro: You’re welcome! – Brian M. Scott Jun 28 '20 at 21:41
  • Hi professor, perhaps I have found an answer to a my question whose nobody answered: could I ask to you if my answer is correct? Click here to see it. Excuse me for the bother. – Antonio Maria Di Mauro Jun 30 '20 at 19:29
  • @AntonioMariaDiMauro: Yes, it’s okay, though I did fix one significant typo early in the proof. I’d prove it differently, without nets, in a way that I think is easier, but what you’ve done works. – Brian M. Scott Jun 30 '20 at 19:41
  • Okay, I'm happy the proof is correct. Anyway if you say you don't use the nets to prove the theorem then I suppose that you define the limit of a function through the filters, right? After all any net determines a filter which converge iff the net converges and limit point is the same for both. – Antonio Maria Di Mauro Jun 30 '20 at 19:51
  • @AntonioMariaDiMauro: No, it can be done just using nbhds. – Brian M. Scott Jun 30 '20 at 19:52
  • Interesting, so how do you proceed? – Antonio Maria Di Mauro Jun 30 '20 at 19:53
  • @AntonioMariaDiMauro: Let $\hat f(x_0)=y_0$ and $\hat f(x)=f(x)$ for $x\in X\setminus{x_0}$. Say that $\lim_{x\to x_0}f(x)=y_0$ iff $\hat f$ is continuous at $x_0$, meaning that for each open nbhd $U$ of $y_0$ there is an open nbhd $V$ of $x_0$ such that $\hat f[V]\subseteq U$. It’s easy to show that if this is true, and $\langle x_n:n\in\Bbb Z^+\rangle\to x_0$ (where $x_n\ne x_0$ for $n\ge 1$), then $\langle f(x_n):n\in\Bbb Z^+\rangle\to y_0$. ... – Brian M. Scott Jun 30 '20 at 20:04
  • ... If it’s not true, $y_0$ has an open nbhd $U$ such that no open nbhd is mapped into it by $\hat f$, so you can take a countable base at $x_0$ and construct a sequence in $X\setminus{x_0}$ converging to $x_0$ whose image doesn’t converge to $y_0$. – Brian M. Scott Jun 30 '20 at 20:04
  • I agree with you your way is more simple. Anyway thanks too much for your assistance! You are very courteous with me! – Antonio Maria Di Mauro Jun 30 '20 at 20:10
  • @AntonioMariaDiMauro: You’re welcome! – Brian M. Scott Jun 30 '20 at 20:12