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I'm trying to determine whether the following space is open with respect to the metric topology

Given the space $M_1=\Bbb R^2$

the distance function $d_1((x,y)(a,b))=|x-a|+|y-b|$

and the subset $V_1\subset \Bbb R^2 := \{(x,y)\in \Bbb R^2 |xy >1\}$

I wanted to find a continous function $f:\Bbb R^2 \rightarrow Y$ where $Z\subset Y$ because of the theorem which states that if Z is open or closed then $f^{-1}(Z)=$V_1$ is open closed.

I have two questions :

1.I want to know how exactly to construct this function because I think I'm doing it wrong...

I think the set is $Z=(1,\infty)$ but what is the function that will map this set to $V_1$.

  1. Is this even the neatest way to prove whether it is open or closed or is there a better method ?
excalibirr
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1 Answers1

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I think you've already constructed such a function ;)

Take $f:\mathbb R^2\to \mathbb R$ to be the function $f:(x,y)\mapsto xy$ and observe that $f$ is by a straightforward analytical argument continuous. It follows that the preimage under $f$ of the open set $(1,\infty)$ is itself an open subset of $\mathbb R^2$, but $f^{-1}(1,\infty)=\{(x,y)\in\mathbb R^2|xy>1\}=V_1$, so $V_1$ is open, as desired.

Edit: To address the second point, it depends entirely on how the set you wish to show open is defined, but in the case of $V_1$ above it's a fairly clean and effective way to go about proving such a thing.

Rafi
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