0

If I define $f:X\rightarrow Y$ to be continuous on a subset $S\subset X$ if for any sequence $\{x_n\}\in S$ such that $x_n\rightarrow x$ implies that $f(x)\in f(S)$.

My question (and it could be a very dumb one, i just can't see why) is: if for any sequence $\{x_n\}\notin S$ such that $x_n\rightarrow x$ it means that $f(x)\notin f(S)$?

If there's an answer to that already, sorry, but i did not find it.

Matheus Silva
  • 75
  • Your definition of continuous on $S$ disagrees with the usual one. For example if $X=Y=\Bbb R$ and $f(x)=x$ for all $x\in \Bbb R$ then according to the usual def'n, $ f$ is continuous on every $S\subset X.$... But if $S=(0,1]$ and $x_n=1/n$ then $x_n\to 0$ but $f(0)=0\not \in (0,1[=f(S).$.... One of the many (equivalent) usual definitions is that $f $ is continuous on $S\subset X$ iff $[;f(x) \in f(S)\cap Cl_Y(f(T))$ whenever $T\subset S$ and $x\in S\cap Cl_X(T);] ,$... where $Cl_X$ and $Cl_Y$ denote the closures in the spaces $X,Y.$ – DanielWainfleet Oct 05 '18 at 21:40
  • The definition of continuity can be fixed by changing $f(x)\in f(S)$ at the end of it to $f(x_n)\to f(x)$. Then presumably the question would be changed so that the desired conclusion is $f(x_n)\not\to f(x)$ instead of $f(x)\notin f(S)$. Alternatively, you could keep $x_n\in S$ but assume $x_n\not\to x$ instead and still ask if $f(x_n)\not\to f(x)$. In both cases, the answer is No, it does not mean that. – Toby Bartels Aug 09 '19 at 19:56

1 Answers1

0

The answer is negative. Take, for instance $f\colon\mathbb{R}\longrightarrow\mathbb{R}$ defined by $f(x)=x$. Then, bu your definition, $f$ is continuous on $[0,1]$. Now, consider the sequence $\left(1+\frac1n\right)_{n\in\mathbb N}$. Its elements don't belong to $[0,1]$. However, it converges to $1$ and $f(1)\in[0,1]$.

José Carlos Santos
  • 427,504