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Let $X,Y$ be topological spaces, and $f:X\to Y$. We say that $f(x)\to l$ as $x \to b$ iff for every open neighborhood $N$ of $l$, there exists an open neighborhood $M$ of $b$ such that $f(M)\subset N$.

There is, however, a different weaker notion of limit: for any sequence $x_n$, whenever $x_n\to b$ as $n\to \infty$, then $f(x_n)\to l$ as $n\to \infty$.

I can prove that the first definition implies the second. In $\mathbb R$, the converse is also true. But is the converse also true in general topological spaces? I don't believe it is true but I cannot find counterexamples.

Paul Frost
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Ma Joad
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  • The weaker notion is known as "sequential continuity". See https://en.wikipedia.org/wiki/Continuous_function#Definition_in_terms_of_limits_of_sequences and https://math.stackexchange.com/q/2343261. – Paul Frost Jun 27 '19 at 08:06
  • BTW your definition needs to be more precise: $b \in X'$ is needed (so $b$ is a limit point of the domain) and we need a truncated neighbourhood of $b$, so $M \setminus {b}$, that maps into $N$. See Wikipedia, e.g. or any good textbook that covers limits like these (Bourbaki, Kelley). – Henno Brandsma Jun 27 '19 at 21:44
  • The co-countable topology is a nice elementary source of a counterexample. Look around on this site. Plenty of hints can be found. – Henno Brandsma Jun 27 '19 at 21:45

1 Answers1

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No.

The standard counterexample is to take $X=\omega_1+1$ with the order topology, $Y$ the two-point discrete space $\{0,1\}$, and $$ f\colon X\to Y;\quad \alpha\mapsto \begin{cases} 1 & \alpha=\omega_1\\ 0 & \text{otherwise} \end{cases}. $$ Then $f$ is sequentially continuous (since the only sequences that converge to $\omega_1$ are eventually $\omega_1$), but not continuous.

user10354138
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