Topology: Show Equivalence of T and T' on R^2

[emob2p]

Homework Statement

I am asked to show that T=[particular point topology on R^2 ((0,0) being the particular point)] is equal to T'=[topology on R^2 from taking the product of R in the particular point topology (0 being particular point) with itself].

The Attempt at a Solution

I'm pretty sure the book is wrong and this problem can't be solved. Consider the set {(0,0),(1,1)}. This is open in T, but I cannot see how this is open in T' since for the point (1,1) we cannot construct an open set that contains (1,1) but is contained in {(0,0),(1,1)}. The closest we can get is {0,1}x{0,1}={(0,0),(1,0),(0,1),(1,1)}. Am I correct, or am I missing something?

 

[HallsofIvy]

Homework Statement

I am asked to show that T=[particular point topology on R^2 ((0,0) being the particular point)] is equal to T'=[topology on R^2 from taking the product of R in the particular point topology (0 being particular point) with itself.

For my own benefit, a set is open in the "particular point topology" if and only if it is either empty of contains the "particular point". In particular, a set in R^2 with the particular point topology, (0,0) being the "particular point", T, if and only if it contains (0,0).
A set is open in the "Cartesian product topology" if and only if it is the Cartesian product of two open sets. Here a set is open in T' if and only if it contains a pair (0,y) and the pair (x, 0).

The Attempt at a Solution

I'm pretty sure the book is wrong and this problem can't be solved. Consider the set {(0,0),(1,1)}. This is open in T, but I cannot see how this is open in T' since for the point (1,1) we cannot construct an open set that contains (1,1) but is contained in {(0,0),(1,1)}. The closest we can get is {0,1}x{0,1}={(0,0),(1,0),(0,1),(1,1)}. Am I correct, or am I missing something?

If I have understood this correctly, yes you are right!

 

[emob2p]

Not quite, the "Cartesian product topology" is generated from the set of Cartesian products of two open sets. If we took your definition for open sets in R^2, then it would be possible for the union of two open sets to not be open. See the difference?