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mathlearner98
mathlearner98
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I know as a fact that $\mathbb{R}$ is the unique (upto isomorphism) complete linearly ordered field. But if we remove the "field" condition and replace it with "dense unbounded set", then I was wondering what is the situation then.

One example I came up was this: Just put $2^{2^\mathbb{N}}-$many copies of $[0,\infty)$ in front of $\mathbb{R}$, and put $2^{2^\mathbb{N}}-$many copies of $(-\infty,0]$ at the back of $\mathbb{R}$ like this:

$\cdots+(-\infty,0]+(-\infty,0]+\mathbb{R}+[0,\infty)+[0,\infty)+\cdots$.

Then this set is dense, unbounded, and complete linearly ordered set (am I right?), and by a cardinality argument, this set isn't isomorphic to $\mathbb{R}$. Please correct me if I am wrong!

I just realized that I wanted to ask some more:

If we instead put $2^\mathbb{N}-$many copies of $[0,\infty)$ and $(-\infty,0]$ at the front and back of $\mathbb{R}$ (instead of $2^{2^{\mathbb{N}}}-$many such copies), then how would the result change?

Also, do you have another example of a complete unbounded dense linearly ordered set that isn't isomorphic to $\mathbb{R}$ in your mind?

I know as a fact that $\mathbb{R}$ is the unique (upto isomorphism) complete linearly ordered field. But if we remove the "field" condition and replace it with "dense unbounded set", then I was wondering what is the situation then.

One example I came up was this: Just put $2^{2^\mathbb{N}}-$many copies of $[0,\infty)$ in front of $\mathbb{R}$, and put $2^{2^\mathbb{N}}-$many copies of $(-\infty,0]$ at the back of $\mathbb{R}$ like this:

$\cdots+(-\infty,0]+(-\infty,0]+\mathbb{R}+[0,\infty)+[0,\infty)+\cdots$.

Then this set is dense, unbounded, and complete linearly ordered set (am I right?), and by a cardinality argument, this set isn't isomorphic to $\mathbb{R}$. Please correct me if I am wrong!

I know as a fact that $\mathbb{R}$ is the unique (upto isomorphism) complete linearly ordered field. But if we remove the "field" condition and replace it with "dense unbounded set", then I was wondering what is the situation then.

One example I came up was this: Just put $2^{2^\mathbb{N}}-$many copies of $[0,\infty)$ in front of $\mathbb{R}$, and put $2^{2^\mathbb{N}}-$many copies of $(-\infty,0]$ at the back of $\mathbb{R}$ like this:

$\cdots+(-\infty,0]+(-\infty,0]+\mathbb{R}+[0,\infty)+[0,\infty)+\cdots$.

Then this set is dense, unbounded, and complete linearly ordered set (am I right?), and by a cardinality argument, this set isn't isomorphic to $\mathbb{R}$. Please correct me if I am wrong!

I just realized that I wanted to ask some more:

If we instead put $2^\mathbb{N}-$many copies of $[0,\infty)$ and $(-\infty,0]$ at the front and back of $\mathbb{R}$ (instead of $2^{2^{\mathbb{N}}}-$many such copies), then how would the result change?

Also, do you have another example of a complete unbounded dense linearly ordered set that isn't isomorphic to $\mathbb{R}$ in your mind?

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mathlearner98
mathlearner98
  • 782
  • 4
  • 14

Example of a complete unbounded dense linearly ordered set that isn't isomorphic to $\mathbb{R}$

I know as a fact that $\mathbb{R}$ is the unique (upto isomorphism) complete linearly ordered field. But if we remove the "field" condition and replace it with "dense unbounded set", then I was wondering what is the situation then.

One example I came up was this: Just put $2^{2^\mathbb{N}}-$many copies of $[0,\infty)$ in front of $\mathbb{R}$, and put $2^{2^\mathbb{N}}-$many copies of $(-\infty,0]$ at the back of $\mathbb{R}$ like this:

$\cdots+(-\infty,0]+(-\infty,0]+\mathbb{R}+[0,\infty)+[0,\infty)+\cdots$.

Then this set is dense, unbounded, and complete linearly ordered set (am I right?), and by a cardinality argument, this set isn't isomorphic to $\mathbb{R}$. Please correct me if I am wrong!