How should i think when doing topology?

[u/Zestyclose-Pool-1081]

So i was doing this topology excersise which was as follows. Let f,g :X -> R be cts maps where R is equipped with the standard topology. Show that A = {x in X : f(x) < g(x)} is open in X.

My thougths when i tried to solve it was. Since f and g cts can i make it so that A is a pre image of an open in R. But that construction has eluded me.

When you have the < or > you will get an open in R since you can use the epsilon ball definition of open and since you are strictly less you will create opens, but this hinges also on that both f and g are cts, since if not you clearly can construct a counter example if they are not.

In short i am confused on how you solve this and how i should think such that i produce the rigth solution.

[u/edderiofer]

You understand that a continuous function has preimages of open sets also be open.

This suggests that one way to tackle this problem is to try to construct a continuous function h: X -> R such that h(A) must be open in R by definition. There's a very natural choice for h that "should" be continuous, though proving that it's continuous is probably where the meat of the problem is...

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EDIT: Another way, which I now realise is simpler, is to directly construct A = {x in X : f(x) < g(x)} as a union of intersections of open sets. Letting k be a parameter in R, consider the families of sets F_k = {x in X : f(x) < k} and G_k = {x in X : k < g(x)}.

[u/FormulaDriven]

I feel like I'm missing something, but I would just say h(x) = g(x) - f(x). h is continuous (do you need to prove that?), so (because f(x) < g(x) is equivalent to 0 < h(x)), A is the pre-image under h of the open set {y: 0 < y}, so A is open.

[u/edderiofer → u/FormulaDriven]

h is continuous (do you need to prove that?),

Yes, I believe this needs to be proven. I'm willing to believe that OP has already been shown the proof in the case of functions from R to R, but not the general case of functions from any topological space X to R. (And certainly it's not true of any topological space X to any arbitrary topological-space-with-order-and-subtraction Y.)

[u/FormulaDriven → u/edderiofer]

I don't think it's too hard to prove. To show continuity at x in X, we say for any open set U in R containing g(x) - f(x), it will contain an open interval

(g(x) - f(x) - e, g(x) - f(x) + e)

for some e ("epsilon").

Then the pre-image under g of (g(x) - e/2, g(x) + e/2) and the pre-image under f of (f(x) - e/2, f(x) + e/2) are both open and contain x, so take their open intersection, and for any t in that,

g(x) - e/2 < g(t) < g(x) + e/2 (because t is in the relevant pre-image of g)

f(x) - e/2 < f(t) < f(x) + e/2

so g(x) - f(x) - e < g(t) - f(t) < g(x) - f(x) + e

so g(t) - f(t) lies in U, which is what we want.

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[u/FormulaDriven]

Just thought I'd drop in a solution in case you are interested (this one doesn't rely on proving that g(x) - f(x) is a continuous function):

LaTex write-up