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How can we prove that the projection map $p\colon\mathbb{R}^2\to\mathbb{R}$, $p(x,y)=x$, is continuous?

This is a very simple question, but I have only had to prove continuity from $\mathbb{R} \to \mathbb{R}$ before, so I would greatly appreciate it if someone could help me how to do this correctly, can we do it using the open balls

Where $c=(x_1,x_2)\in \mathbb{R^{2}}$

$B_{\delta}(c)=\{(x,y): \| (x,y)-c \|<\delta\}$.

$B_{\epsilon}(p(c))=\{x:|x-p(c)|<\epsilon\}$.

And show $\forall c \in \mathbb{R^{2}}$ if $a\in B_{\delta}(c) \implies p(a)\in B_{\epsilon}(p(c)) $.

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    $\begingroup$ One suggestion that works here is that you can rewrite the condition by substituting the definition of $p$. For e.g., say $c = (x_1, x_2)$ and $a = (x,y)$. Then you want to say $$ (x,y) \in B_\delta(x_1, x_2) \implies x \in B_\epsilon(x_1). $$ Your task is: given an $\epsilon > 0$, find some $\delta > 0$ such that the above condition holds. Does this help? $\endgroup$
    – Srivatsan
    Sep 22 '11 at 13:51
  • $\begingroup$ @SrivatsanNarayanan: Indeed, that does help, thanks $\endgroup$
    – Freeman
    Sep 22 '11 at 13:53
  • $\begingroup$ Have you proven the equivalence (in $\mathbb{R}^n$ or "better") of sequential continuity and the ε-δ definition? If so it should be easy: Let $(x_n,y_n)$ be any sequence converging to $(x,y)\ldots$ $\endgroup$
    – kahen
    Sep 22 '11 at 14:19
  • $\begingroup$ @kahen: Have not yet encountered that, I am reading ahead in the course, so my knowledge might be a little bit sketchy. but I have encountered similar things before, if I understand you correctly that looks like a neat way of doing it! $\endgroup$
    – Freeman
    Sep 22 '11 at 14:26
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Let $c=(x_1,x_2)$, and let $\epsilon\gt 0$. You want to show that there exists $\delta\gt 0$ such that if $\lVert (x,y)-(x_1,x_2)\rVert \lt \delta$, then $|x-x_1|\lt\epsilon$.

Let $\delta=\epsilon$. Since $$\lVert (x,y)-(x_1,y_1)\rVert = \sqrt{(x-x_1)^2 + (y-y_1)^2} \geq \sqrt{(x-x_1)^2} = |x-x_1|,$$ then if $\lVert(x,y)-(x_1,y_1)\rVert\lt\epsilon$, then $|x-x_1|\lt\epsilon$. Therefore, $p$ is continuous.

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    $\begingroup$ Excellent! That's what I wrote down immediately after posting this.. good to know it was right :) $\endgroup$
    – Freeman
    Sep 22 '11 at 13:53
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I was going to give Arturo's answer, but I'll just mention that if we endow $\mathbb{R}^2=\mathbb{R}\times\mathbb{R}$ with the product topology, then the projection map $p:\mathbb{R}^2\to\mathbb{R}$ onto the first coordinate is continuous essentially by definition.

One can prove that the product topology on $\mathbb{R}^2$ is the same as the metric topology induced by the metric $$d((x_1,y_1),(x_2,y_2))=\|(x_2-x_1,y_2-y_1)\|=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2},$$ so that whatever answer we get by thinking about $\delta$'s and $\epsilon$'s will be the same as the answer we get by thinking about open sets.

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  • $\begingroup$ Thanks! It's nice to have some more understanding of this in a topological sense $\endgroup$
    – Freeman
    Sep 22 '11 at 14:00
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To add on Zev and Arturo's answers, one can just verify that the preimage of an interval is an open subset of the plane.

In particular, $p^{-1}[(a,b)]=\{\langle x,y\rangle\mid a<x<b, y\in\mathbb R\}=(a,b)\times\mathbb R$, which is an open set in the product.

Therefore the preimage of a basic open set in $\mathbb R$ is an open set in $\mathbb R\times\mathbb R$.

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