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Let $p: (x,y) \subset \mathbb{R}^2 \to x \subset \mathbb{R}$ be the projection function

How do I show that $p$ is continuous?

I considered using the topological definition. Let $A \subset \mathbb{R}$ be open, then $p^{-1}(A)$ returns some set in $\mathbb{R}^2$. We can guarantee that the preimage on the $x$-axis is open, but how can we know that the stuff on the $y$ axis is open? And their cartesian product is open?

Can somebody help? And what is the standard notation for projection, thanks.

Olórin
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1 Answers1

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Let $U\subset \mathbb{R}$ open. Then $p^{-1}(U)=U\times \mathbb{R}$ and hence is open in the box topology (which is equivalent to the standard Euclidean metric).

Using the $\epsilon-\delta$ definition of continuity is also relatively easy, since $p(x,y)=x$, so $|p(x,y)-p(x',y')|=|x-x'|$, so it easy to bound this based on $|(x,y)-(x',y')|$.

Alternatively, if $(x_n,y_n)\to(x,y)$, then $p(x_n,y_n)=x_n\to x=p(x,y)$ so $p$ is continuous.

Moya
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  • Can you clarify what it means to be open on the "box topology"? Is it necessary to know this? – Olórin Mar 20 '16 at 06:30
  • The box topology (or product topology, they're only different for infinite products) on a product of topological spaces, $X\times Y$, is generated by the (sub)basis $U\times V$, where $U\subset X, V\subset Y$ are open. By typical arguments, a function is continuous iff it is continuous on (sub)basis elements. If you're learning topology, then yes, this is very important. – Moya Mar 20 '16 at 06:32