弱于sequential continuous的条件 推出连续性?

hbghlyj

度量空间$(X,d),(Y,e)$, $a\in X$, 函数$f:X\to Y$,
对每个趋于$a$的数列$\an_{n=1}^\infty$存在子列$\{a_{n_k}\}_{k=1}^\infty$使得$f(a_{n_k})\to f(a)$, 则$f$在$a$连续吗?

相关 连续性的标准定义

hbghlyj

与a Continuous Mapping is Sequentially Continuous类似可证.
$\newcommand{\map}[2]{#1 \left( #2 \right)}\newcommand{\sequence}[1]{\left\langle #1 \right\rangle}$
If $f$ is not continuous at $x$, there exists $\epsilon_0 > 0$ such that for all $\delta > 0$ there exists $y \in X$ such that $\map d {x, y} < \delta$ and $\map e {\map f x, \map f y} \ge \epsilon_0$.

For $n \ge 1$, define $\delta_n = \dfrac 1 n$.

For $n \ge 1$, we may choose $y_n \in X$ such that $\map d {x, y_n} < \delta_n$ and $\map e {\map f x, \map f {y_n} } \ge \epsilon_0$.

Therefore, by definition the sequence $\sequence {y_n}_{n \mathop \ge 1}$ converges to $x$.

However, the sequence $\sequence {\map f {y_n} }_{n \mathop \ge 1}$ does not have a subsequence converging to $\map f x$.

Czhang271828

列连续推出连续, 要求 domain 是 $C_1$ (可数邻域基)就行了. 这里度量的条件强太多了.

hbghlyj

Czhang271828 发表于 2023-6-11 06:46
列连续推出连续, 要求 domain 是 $C_1$ (可数邻域基)就行了. 这里度量的条件强太多了. ...

查到了：Neighbourhood basis
For any point $x$ in a metric space, the sequence of open balls around $x$ with radius $1 / n$ form a countable neighbourhood basis $\displaystyle {\mathcal {B}}=\left\{B_{1/n}:n=1,2,3,\dots \right\}$. This means every metric space is first-countable.