Example of a Quotient Map That Is Neither Open Nor Closed

[jmjlt88]

We are just looking for an example of a quotient map that is not open nor closed. Let π: ℝxℝ -> ℝ be a projection onto the first coordinate. Let A be the subspace of ℝxℝ consisting of all points (x,y) such that x≥0 or y=0 or both. Let q:A -> ℝ be a restriction of π. ( Note: assume that q was already proved to be, in fact, a quotient map.) The set ℝx{0} is a subset of A and it is closed in ℝxℝ. Since ℝx{0} is the intersection of A with a closed set of ℝxℝ, it is closed in A. Then, q (ℝx{0}) = ℝ, which is open. Hence, q is not a closed map. Since [0,∞)x (-∞,0) is a subset of A, and it is the intersection of A with ℝx(-∞,0), an open set in ℝxℝ, [0,∞)x (-∞,0) is open in A. However, q ([0,∞)x (-∞,0)) = [0,∞), which is closed. Hence, q is not an open map.

Is this close?

Thank you! :shy:

 

[Dick]

We are just looking for an example of a quotient map that is not open nor closed. Let π: ℝxℝ -> ℝ be a projection onto the first coordinate. Let A be the subspace of ℝxℝ consisting of all points (x,y) such that x≥0 or y=0 or both. Let q:A -> ℝ be a restriction of π. ( Note: assume that q was already proved to be, in fact, a quotient map.) The set ℝx{0} is a subset of A and it is closed in ℝxℝ. Since ℝx{0} is the intersection of A with a closed set of ℝxℝ, it is closed in A. Then, q (ℝx{0}) = ℝ, which is open. Hence, q is not a closed map. Since [0,∞)x (-∞,0) is a subset of A, and it is the intersection of A with ℝx(-∞,0), an open set in ℝxℝ, [0,∞)x (-∞,0) is open in A. However, q ([0,∞)x (-∞,0)) = [0,∞), which is closed. Hence, q is not an open map.

Is this close?

Thank you! :shy:

R is both open and closed in R, so the first part doesn't work. Think about the graph G of y=1/x for x>0. Is G open or closed? What about q(G)?