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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 $A$.

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I edited your "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 meta.math.stackexchange.com/questions/1803/… –  user31373 Sep 9 '12 at 1:37

3 Answers 3

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

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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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First 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? –  drew Sep 8 '12 at 20:01

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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Thanks for the help on how to edit and ask questions. Now heres my attempt at a proof. –  drew Sep 9 '12 at 2:52
    
PROOF: Let ${\g_s : s \in S \}$ be a chain in $A$ and $g$ is a function such that D_g=\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 –  drew Sep 9 '12 at 3:10
    
@drew It would be better to post this as an answer than a comment: you are not commenting about my answer, but writing your own. Also, I can't help you with formatting here since I can't edit comments. –  user31373 Sep 9 '12 at 3:16
    
Sorry I am new to this. But it should be fixed now. –  drew Sep 9 '12 at 3:26

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