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This is a problem from an old homework assignment which I never got, and I've decided to go back and try again. As a side note, I'm quite new to writing proofs.

Definition: $\alpha \subseteq \mathbb{R}$ is ideal if

  1. $\forall x \in \alpha, \forall y \in \mathbb{R}, xy \in \alpha$ and
  2. $\forall x,y \in \alpha, x+y \in \alpha$

I've split up the claim as such: if $\alpha \subseteq \mathbb{R}$ is ideal then

  • $\alpha \neq \{0\} \implies \alpha = \mathbb{R}$ and
  • $\alpha \neq \mathbb{R} \implies \alpha = \{0\}$

Proof:

(first bullet)

($\subseteq$) this is given.

($\supseteq$) Let $a \in \mathbb{R}$. Since $\alpha \neq \{0\}$, $\exists b \in \alpha$ such that $b \neq 0$. Combining (1) and (2), we get

$$b+ab = c$$

for some $c \in \alpha$. Rearranging this equation, we get

$$a+1=\frac{c}{b}$$

(recall that $b \neq 0)$. Since $\alpha \subseteq \mathbb{R}$, we have that $b \in \mathbb{R}$, and we also know that $\frac{1}{b} \in \mathbb{R}$. Thus, by (1),

$$\frac{c}{b} = c \cdot \frac{1}{b} = a+1 \in \alpha$$

But $a+1$ is still an arbitrary element in $\mathbb{R}$, so we have shown that $\mathbb{R} \subseteq \alpha$, as needed.

(second bullet)

($\supseteq$) Let $a \in \alpha$. Then, by (1), $a \cdot 0 = 0 \in \alpha$, as needed.

($\subseteq$) Let $a \in \alpha$. Since $\alpha \neq \mathbb{R}$, $\exists b \in \mathbb{R}$ such that $b \notin \alpha$. By combining (1) and (2), we get

$$a + ab = a(1+b) \in \alpha$$


This is where I'm stuck. I know that I need to somehow show that $a$ must be equal to $0$, but I can't figure out where to go from here. Any hints? Also, is my proof of the first bullet valid?

Thanks!

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  • $\begingroup$ The two claims that you've written are logically equivalent to each other. If you prove any one, you've automatically proved the other. A statement of the form $P \implies Q$ is equivalent to the statement $\neg Q \implies \neg P$ (where $\neg$ implies negation, or "not"). The latter is statement is called the "contrapositive" (of the former). So, if you prove that $\alpha \ne \{0\} \implies \alpha = \mathbb R$, then of course, if $\alpha \ne \mathbb R$, then it must be that $\alpha = \{0\}$. $\endgroup$ – M. Vinay Mar 21 at 1:49
  • $\begingroup$ Assume that $\alpha\neq \mathbb{R}$. If $a\in\alpha$ and $a\neq0$ then there is $a^{-1}\in\mathbb{R}$ such that $a^{-1}a=1$. Since $\alpha$ is ideal, then for all $r\in\mathbb{R}$ you hav ethat $r=r1=(ra^{-1})a\in\alpha$. Contradiction. $\endgroup$ – user647486 Mar 21 at 1:50
  • $\begingroup$ Your proof of the first bullet is correct, well done! Let me suggest a simplification though: You don't need to use both 1 and 2 in that step. From 1 itself, you know that $ab = c \in \alpha$, and since $b \ne 0$, you get $a = \frac c b = \frac 1 b \times c \in \alpha$. $\endgroup$ – M. Vinay Mar 21 at 1:55
  • $\begingroup$ Thanks @M.Vinay! Very helpful points. $\endgroup$ – Archie Gertsman Mar 21 at 2:04
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For the first bullet, I think you could simplify by noting that since $b \in \alpha$ with $b\neq 0$, we have $$1 = \frac{1}{b} \cdot b \in \alpha$$ as $\frac{1}{b} \in \mathbb{R}$. This implies $\alpha = \mathbb{R}$ by (1).

For the second bullet point, we assume $\alpha \neq \mathbb{R}$ and so $\exists x \in \mathbb{R}$ with $x \not\in \alpha$. This implies $1 \not\in \alpha$, else we would have $x \cdot 1 \in \alpha$. This implies $\alpha \subset \{0\}$ as otherwise $a \in \alpha$ with $a\neq 0 \implies 1 \in \alpha$ as shown above.

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I'd attack the problem a little differently. Assume $0 \neq x \in \alpha$ and choose an arbitrary $z \in \Bbb R$. Then by property $1, x\frac{z}{x} = z \in \alpha$. ($\frac{z}{x}$ exists because $x \neq 0$.) But $z$ was an arbitrary element of $\Bbb R$, so $\exists x \in \alpha, x \neq 0 \Rightarrow \alpha = \Bbb R$.

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