Showing a set is closed with the definition of continuity

[radou]

Homework Statement

I need to show that the subset of R^2 given with A = {(x, y) : xy = 1} is closed by using the "closed set formulation" of continuity.

The Attempt at a Solution

So, if a function f : X --> Y is continuous, then for every closed subset B of Y, its preimage f^-1(B) is closed.

The set A can be written as A = {(x, 1/x) : x is in R\{0}}. Since f(x) = 1/x is a continuous function on R\{0}, and since any subset of R containing only one element is closed, f^-1{a} = {1/a} is a closed subset of R, for every a in R\{0}. Any ordered pair of the form (a, 1/a) can be written as a cartesian product of the sets {a} x {1/a}, which is closed, since the sets are closed. But an infinite union of such sets need not be closed. I feel it's warm around here, but I just can't figure it out.

Perhaps I'm not on the right track. Any help appreciated, as always.

 

[radou]

...no thoughts?

 

[HallsofIvy]

One major difficulty with your proof is that your function is NOT continuous on any set including x= 0! And there is nothing in your initial statement that says you can exclude x= 0.

I would be inclined, instead, to use the function f(x, y)= xy. That is continuous for all (x, y) in R².

 

[radou]

I would be inclined, instead, to use the function f(x, y)= xy. That is continuous for all (x, y) in R².

Thanks a lot. In this case, the proof is almost trivial - since {1} is a closed set in R, its preimage must be closed, and it is exactly the set we're looking at.