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Let $\mathbb{R}_{\tau}$ be the set of real numbers with topology $\tau = \{(-x,x)| x>0\} \cup \{\emptyset, \mathbb{R}\}$ and $\mathbb{R}_{\tau} \times \mathbb{R}_{\tau}$ be the product topology on $\mathbb{R}^2$.

a) Prove that $A = \{(x,y) \in \mathbb{R}^2 | x^2 + y^2 < 1\}$ is open in $\mathbb{R}_{\tau} \times \mathbb{R}_{\tau}$.

b) Find $\overline{A}$. Justify your answer.

c) What functions $f: {\mathbb{R}_{\tau}}^2 \rightarrow \mathbb{R}$ are continuous? Here $\mathbb{R}$ has the standard topology and ${\mathbb{R}_{\tau}}^2 = \mathbb{R}_{\tau} \times \mathbb{R}_{\tau}$ has the product topology.

Please help!

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1  
What have you tried? –  Chris Eagle Oct 16 '12 at 20:49
    
got stuck here. –  John Lennon Oct 16 '12 at 21:01
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You got stuck before starting? –  Chris Eagle Oct 16 '12 at 21:04
    
need to know what to start with –  John Lennon Oct 16 '12 at 21:08
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looking up the definitions and trying to apply them.. –  Berci Oct 16 '12 at 21:08

1 Answer 1

Hints:

a) You can write up $A$ as a union of (many) open sets. You need to know how the product topology is defined, and then consider all open filled rectangle which are centered at the origo and are inside the given disk $A$. For any $(x,y)\in A$, show such an open rectangle that contains $(x,y)$.

b) For this consider the closed filled rectangles (still centered in the origo), which include $A$. These are all going to be closed (why?), and take their intersection.

c) The open intervals form a basis for the standard topology (i.e. every open set is the union of them), and then the requirement that $f$ is continuous becomes to verifying that $f^{-1}(I)$ is open in $\Bbb {R_\tau}^2$, i.e. union of open filled rectangles centered at origo, for every open interval $I=(a,b)$..

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