Question: I want to show the map $f:\mathbb R^2\rightarrow\mathbb R$, defined as $f(x,y)=x+y$ is continuous for all $(x,y)\in\mathbb R^2$.

Issue: I know how to prove this via the epsilon-delta way. I want to prove this using the projection functions $p_1,p_2: \mathbb R^2\rightarrow\mathbb R$ where $p_1$ maps $(x,y)\rightarrow x$, similarly for $p_2$. Now my books says via a result based on continuous functions from $\mathbb R\rightarrow\mathbb R$ that the sum of $p_1+p_2$ is continuous but I don't see how that would be possible as the domain of the functions are $\mathbb R$ and not $\mathbb R^2$.

  • 1
    $\begingroup$ It is a polynomial function: how continuous do you want it to be? It is also the sum of two projections, which are continuous in each variable (and here you can widen the argument from $\;\Bbb R\;$ . $\endgroup$ – Timbuc Dec 20 '14 at 1:59
  • 1
    $\begingroup$ @Timbuc I never questioned whether it was continuous. My book is showing me an alternative way which I don't logically agree with. I accept the projection functions are continuous $\forall(x,y) \in \mathbb R^2$ but I don't see how you can say the sum of the two projections is continuous based on results from real to real functions (i.e. if $f,g$ are continuous functions from $\mathbb R$ to $\mathbb R$ then so is $f+g$, I don't understand how they have applied this...) $\endgroup$ – Raul Dec 20 '14 at 2:05
  • $\begingroup$ @MPW Is there a way to prove the question using the projection functions, but only using results based on continuous real-valued functions from the reals. $\endgroup$ – Raul Dec 20 '14 at 2:14
  • $\begingroup$ If you concede that addition $s:\mathbb R\times \mathbb R\rightarrow R$ (with $s(x+y)=x+y$) is continuous, then $f=s\circ (p_1\times p_2)$ is continuous. $\endgroup$ – MPW Dec 20 '14 at 2:15
  • 1
    $\begingroup$ @MPW so there isn't really a clear cut way then....you have gone from the results I want to show and forcefully got $p_1,p_2$ involved it seems. $\endgroup$ – Raul Dec 20 '14 at 2:21

Your Answer

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

Browse other questions tagged or ask your own question.