Let $A$ be the set of all functions $f$ with domain $D_f$ a subset of $X$ and range $R_f$ a subset of $Y$ and let $A$ be a partially ordered by extension; that is $g\preceq f$ if and only if $g\subseteq f$. If $\{g_s : s\in S\}$ is a chain in $A$, show that the function $g$ such that $D_g$ equals $\bigcup_{s\in S} D_{g_s}$, and $g(x)=g_s(x)$ for all $x\in D_{g_s}$ is an upper bound for this chain in $A$.
Real Analysis Least upper bound of a set of functions?
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1I edited you$r$ "question" using LaTeX. If you click *edit* on it, you can see exactly how this is done. Also note that "questions" which are verbatim copies of textbook/homework problems are considered low-quality material, and are voted on accordingly. See http://meta.math.stackexchange.com/questions/1803/how-to-ask-a-homework-question – 2012-09-09
3 Answers
PROOF: Let $\{g_s : s\in S \}$ be a chain in $A$ and $g$ is a function such that $D_g$ equals $\bigcup_{s\in S} D_{g_s}$ and $g(x)=g_s(x)$ for all $x\in D_{g_s}$. Assume to the contrary there is a function $g_t$ where $t\in S$ that is not a subset of $g$ but $D_{g_t}$ is a subset of $D_g$ if $x\in D_{g_t}$ then $x\in D_g$ by definition $g_t(x)=g(x)$ for all $x\in D_{g_s}$. Hence $g_t\subset g$ by contradiction $g$ is an upper bound of $\{$g_s$ : s\in S$}. QED
First we show that $g$ is a function. Let $x \in D^g = \bigcup_{s\in S} D^{g^s}$, if there are $s, t\in S$ with $x \in D^{g^s} \cap D^{g^t}$, we have wlog $g^s \subseteq g^t$ and hence $g^s(x) = g^t(x)$. So $g(x)$ is well-defined.
Now let $s \in S$ and $x \in D^{g^s}$, then, by definition $x \in D^g$ and $g(x) = g^s(x)$. So $g^s \subseteq g$. As $s$ was arbitray, $g$ is an upper bound of $\{g^s \mid s \in S \}$.
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0First off thanks for your help. When you used wlog because gs is a chain either gs precedes gt or gt precedes gs and since the arguments are similar because all you have to do is switch the s and t you only have to show one way? Also you can conclude x is in Dg because if x is in Dgs and Dg is equal to the union of all the Dgs then x must be in Dg, so therefore g(x)=gs(x) because u previously showed g was well defined. Then gs is a subset of g because the g(Dg)=g1(Dg1) U g2(Dg2) U...U gs(Dgs) since you only picked one s out of S gs is a subset of g? – 2012-09-08
A function is a subset of $X\times Y$ that passes the vertical line test. You are given a chain of such subsets. The union is an upper bound since it contains each such subset. Why does the union pass the vertical line test? Suppose it fails; some vertical line has at least two common points with it. These two points belong to some $g^s$ and $g^t$, hence both belong to $g^{\max (s,t)}$. Contradiction.
When writing up your solution, be sure to replace the informal term "vertical line" with its formal equivalent.
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0Sorry I am new to t$h$is. But it should be fixed now. – 2012-09-09