The definition says: Let $(X,d_{1})$ and $(Y,d_{2})$ be a metric space. A map $f:X\rightarrow Y$ is called continuous if for every $x\in X$ and $\epsilon>0$ there exists a $\delta>0$ such that:

$d_{1}(x,y)<\delta\Rightarrow d_{2}(f(x),f(y))<\epsilon$

does it mean that $x,y\in X$ and $f(x),f(y)\in Y$?

  • $\begingroup$ Yes, that's exactly what that means. $\endgroup$ – DMcMor May 4 '17 at 20:12
  • $\begingroup$ @DMcMor ok, so the use of elements $x,y$ was confusing, thanks! $\endgroup$ – gbox May 4 '17 at 20:13

Your wording is a bit unclear but here's a rephrasing of the definition that may be a bit more clear:

Fix $x\in X$. We say that $f:X\to Y$ is continuous at $x$ if for every $\epsilon>0$, there exists $\delta>0$ such that the following condition holds: if $y\in X$ such that $d_1(x,y)<\delta$, then $d_2(f(x),f(y))<\epsilon$.

We say $f$ is continuous if it is continuous at every $x\in X$.

If I've interpreted your question wrong, let me know and I can try to update my answer.


$\;\;\;\;f $ is continuous at $X $ $$\iff $$

$(\forall x\in X )\;\;\;f $ is continuous at $x $ $$\iff $$ $$(\forall x\in X) \;\;(\forall \epsilon>0)\;\;(\exists \delta>0 )\;:\;\color {green}{(\forall y\in X )}$$ $$(d_1 (x,y)<\delta\implies d_2 (f (x),f (y))<\epsilon).$$

of course $f (x) $ and $f (y) $ are in $Y $ since $f :X\to Y $.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.