Proof: Let $\mathcal{F}$ be the set of all subrings of $\mathbb{R}$ not containing $\dfrac{1}{2}$, and let $\mathcal{C}$ be a typical chain of subrings $R_0 \subseteq R_1 \subseteq ...$. Admit$$R=\bigcup_{n∈N}R_n$$and prove it is an upper bound of $\mathcal{C}$. For any $a,b∈R$, we have $a∈R_x$ and $b∈R_y$ for some $x,y$ implying $a-b∈R$ and $ab∈R$ so that $R$ is a subring. Furthermore, by definition of the union, $1∈R$ and $\dfrac{1}{2} \not ∈ R$. By Zorn's Lemma, $\mathcal{F}$ has a maximal element.$~\square$
Thursday, May 2, 2013
Zorn's Lemma on the Real Numbers (7.5.6)
Dummit and Foote Abstract Algebra, section 7.5, exercise 6:
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Prove that $\mathbb{R}$ contains a subring $A$ with identity and maximal under inclusion such that $\dfrac{1}{2} \not ∈ A$.
Proof: Let $\mathcal{F}$ be the set of all subrings of $\mathbb{R}$ not containing $\dfrac{1}{2}$, and let $\mathcal{C}$ be a typical chain of subrings $R_0 \subseteq R_1 \subseteq ...$. Admit$$R=\bigcup_{n∈N}R_n$$and prove it is an upper bound of $\mathcal{C}$. For any $a,b∈R$, we have $a∈R_x$ and $b∈R_y$ for some $x,y$ implying $a-b∈R$ and $ab∈R$ so that $R$ is a subring. Furthermore, by definition of the union, $1∈R$ and $\dfrac{1}{2} \not ∈ R$. By Zorn's Lemma, $\mathcal{F}$ has a maximal element.$~\square$
Proof: Let $\mathcal{F}$ be the set of all subrings of $\mathbb{R}$ not containing $\dfrac{1}{2}$, and let $\mathcal{C}$ be a typical chain of subrings $R_0 \subseteq R_1 \subseteq ...$. Admit$$R=\bigcup_{n∈N}R_n$$and prove it is an upper bound of $\mathcal{C}$. For any $a,b∈R$, we have $a∈R_x$ and $b∈R_y$ for some $x,y$ implying $a-b∈R$ and $ab∈R$ so that $R$ is a subring. Furthermore, by definition of the union, $1∈R$ and $\dfrac{1}{2} \not ∈ R$. By Zorn's Lemma, $\mathcal{F}$ has a maximal element.$~\square$
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