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$\def\Rng{\operatorname{Rng}}$ Let $R$ be a relation from $A$ to $B$. For $a \in A$, define the vertical section of $R$ at $a$ to be $R_a$ = $\{ y \in B: (a,y) \in R\}$. Prove that the union over $R_a$ where $a \in A = \Rng(R)$.

I have work for it but as you can see I have no formatting abilities, sorry! I worked through the definitions of $R_a$ and $\Rng(R)$ and came to the proper conclusions by showing inclusion in both directions but it seems iffy to me.

Przemysław Scherwentke
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Wes
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1 Answers1

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So if $y \in \bigcup_{a \in A} R_a \Rightarrow y \in R_x$ for some $x \in A$, then $(x, y) \in R$, hence $y \in \operatorname{Rng}(R)$. Conversely $y \in \operatorname{Rng}(R)$ implies that there exists $x \in A$, such that $(x, y) \in R \Rightarrow y \in R_x \subseteq \bigcup_{a\in A} R_a$.

Przemysław Scherwentke
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brick
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  • What justifies that y an element of R_x implies y an element of the union over R_a? – Wes Oct 30 '14 at 23:44
  • @WesleyJordt Well it's because $x \in A$ and $y \in R_x$. The union contains all elements $y$ such that $y \in R_x$ for some $x \in A$. – brick Oct 30 '14 at 23:51