- #1

issacnewton

- 1,026

- 36

- Homework Statement
- Suppose that the non-empty subset ##A## of ##\mathbb{R}## is bounded above. Then ##A##

has a least upper bound in ##\mathbb{R}## i.e. lub ##A## exists.

- Relevant Equations
- definition of Dedekind cut

So, I have to come up with some set which is lub A. Now, A is a subset of R, so each member of A is a Dedekind left set. So, A is a set of sets. Now, I propose that the following set would be lub of ##A##.

$$ \alpha = \bigcup A = \{ \beta | \exists \delta \in A (\beta \in \delta) \} $$

Actually, I have proven most of it. I am at the part, where I am trying to prove that ##\alpha## is a real number. So, for that, I will need to prove that, ##\alpha## is a Dedekind left set. So, I need to prove that ##\alpha \ne \mathbb{Q} ##. Which means that, there exists some ##x \in \mathbb{Q} ## such that ## x \notin \alpha##. This is what I need to prove. Now, I am given that ##A## is bounded above, so there is some ##M \in \mathbb{R} ##, such that

$$ \forall \; x \in A \; (x \leqslant M) \cdots\cdots (1) $$

Now, ##M < M+1##. There is some ##y \in \mathbb{Q} ## such that ## M < y < M + 1##. So, it means that

$$ \forall \; x \in A \; (x < y) \cdots\cdots (2 )$$

Now, assume the negation that ## y \in \alpha ##. Using the definition of ##\alpha##, it means that there is some ##N \in A## such that ##y \in N##. Since ##N \in A##, using equation 2, it follows that ## N < y##. But ##N## and ##y##, are sets. So, this means that, ##N## is a proper subset of ##y##. ##N \subset y ##. Since, I have ## y \in N##, it follows that ## y \in y##. And here is a contradiction according to the ZF set theory. So, my assumption that ## y \in \alpha ## is wrong. So, I have ## y \notin \alpha ##. This means that ##\exists y \in \mathbb{Q}## such that ## y \notin \alpha##. This proves that ##\alpha \ne \mathbb{Q}##.

Is the proof correct ? I have proven other properties here.

$$ \alpha = \bigcup A = \{ \beta | \exists \delta \in A (\beta \in \delta) \} $$

Actually, I have proven most of it. I am at the part, where I am trying to prove that ##\alpha## is a real number. So, for that, I will need to prove that, ##\alpha## is a Dedekind left set. So, I need to prove that ##\alpha \ne \mathbb{Q} ##. Which means that, there exists some ##x \in \mathbb{Q} ## such that ## x \notin \alpha##. This is what I need to prove. Now, I am given that ##A## is bounded above, so there is some ##M \in \mathbb{R} ##, such that

$$ \forall \; x \in A \; (x \leqslant M) \cdots\cdots (1) $$

Now, ##M < M+1##. There is some ##y \in \mathbb{Q} ## such that ## M < y < M + 1##. So, it means that

$$ \forall \; x \in A \; (x < y) \cdots\cdots (2 )$$

Now, assume the negation that ## y \in \alpha ##. Using the definition of ##\alpha##, it means that there is some ##N \in A## such that ##y \in N##. Since ##N \in A##, using equation 2, it follows that ## N < y##. But ##N## and ##y##, are sets. So, this means that, ##N## is a proper subset of ##y##. ##N \subset y ##. Since, I have ## y \in N##, it follows that ## y \in y##. And here is a contradiction according to the ZF set theory. So, my assumption that ## y \in \alpha ## is wrong. So, I have ## y \notin \alpha ##. This means that ##\exists y \in \mathbb{Q}## such that ## y \notin \alpha##. This proves that ##\alpha \ne \mathbb{Q}##.

Is the proof correct ? I have proven other properties here.