0

When we have two topological spaces, $(X, \tau_X)$ and $(Y, \tau_Y)$ it is easy to check that for $f: X \rightarrow Y$ continuity implies sequential continuity. I'm wondering what do we have to assume for $X$, $Y$ for the converse to hold: here, sequential continuity vs. continuity, are given two sufficient properties ($X$ being sequential, $X$ being first countable). Can it be switched with something else? In particular I'm interested if it holds when $X$ is locally compact and Hausdorff, $Y=\mathbb{K}\in \{ \mathbb{R},\mathbb{C}\}$.

I'd like just a hint, not full proof.

Another User
  • 5,048
Mogget
  • 81
  • I don't think so. Sequential spaces are pretty much defined to be the weakest spaces where such a condition holds. $[0,1]^{[0,1]}$ in the product topology is compact and Hausdorff, but is not sequential. See this: http://repository.cmu.edu/cgi/viewcontent.cgi?article=1164&context=math – Moya Apr 01 '15 at 04:10
  • Thank You. I don't really see why is it sequential though. I've tried to show that $\left( \prod_{t \in [0,1]}]\frac{1}{2},1]\right)' $ is sequentially closed but it doesn't seem to be the case. – Mogget Apr 02 '15 at 12:16

0 Answers0