Intersection of sets in ##\Bbb{R}^2## and Open Maps

[Bashyboy]

Homework Statement

No problem statement

Homework Equations

The Attempt at a Solution

Let $A = \{(x,y) ~|~ x \ge 0 \mbox{ or } y=0 \}$ be a subspace, which can be shown closed, of $\Bbb{R}^2$. If my calculations are right, isn't $A \cap [(0,\infty) \times \Bbb{R} ] = (\{0\} \times \Bbb{R}) \cup ((0,\infty) \times \{0\})$ and therefore $\pi_1(A \cap [(0,\infty) \times \Bbb{R} ] ) = [0,\infty)$, where $\pi_1 : \Bbb{R}^2 \to \Bbb{R}$ the canonical projection map onto the first coordinate, thereby showing that $\pi_1$ restricted to $A$ is not an open map?

 

[fresh_42]

$(0,0)$ isn't an element of your second set $(0,\infty) \times \Bbb{R}$.

$A \cap [(0,\infty) \times \Bbb{R} ] = (\{0\} \times \Bbb{R}) \cup ((0,\infty) \times \{0\})$

 

[LCKurtz]

Homework Statement

No problem statement

Homework Equations

The Attempt at a Solution

Let $A = \{(x,y) ~|~ x \ge 0 \mbox{ or } y=0 \}$ be a subspace,

But it isn't a subspace.

 

[fresh_42]

Why not, isn't $0$ in $\mathbb R$?

Yes, but why is it in $(0,\infty)$?

 

[WWGD]

Yes, but why is it in $(0,\infty)$?

Never mind, let me delete. I was somehow thinking of intervals. The notation $(a,b)$ used both for points and intervals can be confusing.

 

[fresh_42]

Never mind, let me delete. I was somehow thinking of intervals. The notation $(a,b)$ used both for points and intervals can be confusing.

Yes, therefore I like $]0,\infty[$ for open intervals a lot more. But I seem to be the only one here.

 

[WWGD]

Yes, therefore I like $]0,\infty[$ for open intervals a lot more. But I seem to be the only one here.

What do you use for closed intervals?

 

[fresh_42]

$[a,b]$ closed, $[a,b[$ half-open and $]a,b[$ open. I wonder where the round ones came from.

 

[Bashyboy]

Ah! You are right. I think I know how to solve the problem now, which is to find a set that is open in $A$ such that its image under $q$ is not open in $\Bbb{R}$. The idea is to intersect the $y$-axis with an open ball. More specifically, consider $A \cap B((0,2),1) = \{(x,y) \in A \mid x^2 + (y-2)^2 < 1 \}$. I claim that the image of this is $[0,1)$. If $(x,y) \in A \cap B((0,2),1)$, then $x \ge 0$ or $y=0$ and $x^2 + (y-2)^2 < 1$, which implies $|x| < 1$. If $x \ge 0$, then $q(x,y) = x \in [0,1)$ (note: the $y=0$ case cannot obtain). Now, if $x \in [0,1)$, then $q(x,2) = x$ where #(x,2) \in A \cap B((0,2),1)##.

How does this sound?