How to prove that this set is closed?

Question

Suppose $Y$ be an ordered set in the order topology. Let $f, g: X \to Y$ be continuous. How to show that the set $\{x| f(x) \leq g(x)\}$ is closed?

This is a excercise from munkres. Maybe trying to show that the set $f(\{x| f(x) > g(x)\})$ is open is suffice. But I could not figure out the connection between this set and the continuity of the two functions.Could you give a hint?

Thanks.

Henno Brandsma · Accepted Answer · 2017-09-19T13:44:28.180

7

The set $$U = \{(y_1, y_2) \in Y \times Y \mid\, y_1 \leq y_2 \,\}$$ is closed in $Y \times Y\,$:

Suppose $(z_1, z_2) \in Y \times Y\,$ is not in it, so that $z_1 > z_2$.

If there is some $z_3 \in Y$ such that $z_1 > z_3 > z_2$, the basic open set $(z_3, \rightarrow) \times (\leftarrow, z_3)$ contains $(z_1, z_2)$ and misses $U$, otherwise $z_1$ and $z_2$ are neighbours and $(z_2, \rightarrow) \times (\leftarrow, z_1)$ has this property. Note that all used subsets in $Y$ are open in the order topology.

Now $f \nabla g: X \rightarrow Y \times Y$ is continuous (standard in the product topology) when $f$ and $g$ are, and your set is exactly

$$(f \times g)^{-1}[U]$$

and thus closed as the inverse image of a closed set under a continuous function.

edited Sep 19 '17 at 13:44

answered Sep 21 '11 at 11:43

Henno Brandsma

242,131

1

Calling a closed set $U$ is strange :-) But I guess it was $U$ for upper half space... – lhf Sep 21 '11 at 12:02
@lhf: it was for "upper" indeed. – Henno Brandsma Sep 21 '11 at 12:07
@HennoBrandsma I have just editted a post of yours from 2011! Do you agree to this edit? I guess you mean the map $f \times g \colon X \rightarrow Y \times Y$ defined by $\left( f \times g \right) (x) \colon= f(x) \times g(x)$ for all $x \in X$. Am I right? – Saaqib Mahmood Sep 19 '17 at 08:30
@SaaqibMahmuud yes, you're almost right. I meant the diagonal product $f \nabla g$ that sends $x$ to $(f(x),g(x))$ of course. – Henno Brandsma Sep 19 '17 at 13:46
@HennoBrandsma thanks. What do you mean by "almost right"? – Saaqib Mahmood Sep 19 '17 at 15:40
But where does the existence of $z_3$ comes from ? – Our Dec 18 '17 at 07:40
@onurcanbektas there are 2 cases: such a $z_3$ exists or not – Henno Brandsma Dec 18 '17 at 16:16
@HennoBrandsma I read your proof in the morning, so being little sleepy led me to not being able to see that "if" part :). Thanks for your response. – Our Dec 18 '17 at 16:22
@SaaqibMahmuud I'm against the confusing notation $f(x) \times g(x)$ for $(f(x), g(x))$. – Henno Brandsma Dec 19 '17 at 12:55

lhf · Answer 2 · 2011-09-21T11:24:49.577

6

The complement of $\{x : f(x) \leq g(x)\}$ is $\{x : f(x) > g(x)\}$. Part of the latter set is $$\bigcup_{y\in Y} \{x : f(x) > y > g(x)\},$$ which is open because each set is the intersection of two open sets. What is missing?

edited Sep 21 '11 at 11:24

answered Sep 21 '11 at 11:22

lhf

216,483

How to prove that this set is closed?

2 Answers2

Linked

Related