[증명] the set {x|f(x)≤g(x)} is closed in X

명제.

Let Y be an ordered set in the order topology. Let f, g : X → Y be continuous. Then the set {x|f(x)≤g(x)} is closed in X.

 

증명.

바로 전에 증명한 명제 하나와 따름정리 하나를 이용할 것이다.

 

I will use the theorem that a simple ordered set is hausdorff (proof: https://wangdo.tistory.com/59) and the corollary in https://wangdo.tistory.com/60.

 

A simple ordered set is hausdorff, so Y is hausdorff.

Then, since Y is hausdorff, by the corollary in https://wangdo.tistory.com/60, W = { x × y | x ≤ y ∈ Y } is closed in Y × Y.

 

Consider the function (f × g) : X → Y × Y, and (f × g)(x)=f(x) × g(x) ∀x∈X.

 

Because f and g are continuous, (f × g) is also continuous, which means the preimage of a closed set is closed in X.

Now, the preimage of W will be the set {x|f(x)≤g(x)}. Here, W is closed so {x|f(x)≤g(x)} is also closed.

 

증명에 이상이 있다면 댓글로 의견 바랍니다.

If there are any errors, please let me know in the comments.