For the following question written below, what I would like to ask is, are the open set from an ordered topology $\mathcal{U}$ of a linearly ordered set $Y.$ The reason I ask is that to answer the question, if I want to use the open set definition for continuity, I have to specify what the open sets are specifically.
Let $(Y,\leq)$ be a linearly ordered set, and let $\mathcal{U}$ denote the order topology on $Y$. Furthermore, let $(X,\mathcal{F})$ be a topological space and let $f,g:X\rightarrow\ Y$ be continuous functions.
Define a function $h:X \rightarrow\ Y$ by $h(x)=\min\{f(x),g(x)\}$ for all $\ x\in\ X$. Prove that $h$ is continuous.
Thank you in advance!
Best Answer
For $y\in Y$ let $(-,y)=\{y'\in Y: y'<y\}$ and $(y,+)=\{y'\in y:y<y'\}.$
Let $L=\{(-,y):y\in Y\}$ and $U=\{(y,+):y \in Y\}.$
Let $B=\{Y\} \cup L \cup U \cup \{l\cap u:l\in L\land u\in U\}.$
Then $B$ is a base for the order-topology on $Y. $
For $y,y'\in Y$ let $(y,y')=\{y''\in Y: y<y''<y'\}.$ This is called a bounded open interval. The members of $\{l\cap u:l\in L\land u\in U\}$ are the bounded open intervals (including the empty set).
Any $A\subset Y$ is convex iff $\forall a,a'\in A\; (\,(a,a')\subset A).$The members of the base $B$ are precisely the convex open sets in the order-topology. So any open set in the order-topology is the union of a family of convex open sets.
This is applicable to all $Y,$ i.e. regardless of whether $Y$ has a max or min or any other specific properties. In particular (as a caution) it is possible for a bounded open interval to have a max or min.
A useful property for any $Y$ is that the closure of $(-,y)$ is $(-,y)\cup \{y\}$ iff $y=\sup (-,y)$ iff $(-,y)$ is not empty and has no max. And $(-,y)$ is closed iff $(-,y)$ is empty or has a largest member. And similar considerations hold for $(y,+).$
From that, we see that if $y\in Y$ and $y\not \in A\subset Y$ then $y\in \overline A$ iff $[\,y=\sup (A\cap (-,y)) \lor y=\inf (A\cap (y,+)\,)\,].$