Prove, using the definition of a closed set, that $S = [0, \Omega)$ is not a closed subset of $X = [0, \Omega]$ with order topology

Question

I am using the book A First Course in Topology by Robert Conover. I can assume everything up to this point and info on ordinals.

Prove using the definition of a closed set that S = [0,$\Omega$) is not a closed subset of $X = [0, \Omega]$ with order topology (prob 4.a, pg 68).

Assumptions and definitions

$\Omega$ is the the initial ordinal of $\text{card}(\aleph_1)$.

https://en.wikipedia.org/wiki/Ordinal_number

5.1 Theorem Any countable subset of $[0, \Omega)$ is bounded above (i.e., if $S \subset[0, \Omega]$ with $\lvert S \rvert \leq \aleph_0$, then there exists a number x $\in$ [0,$\Omega$] such that $x \geq s$ for every $s \in S$).

My attempt(edit)

LetS, X=[0,$\Omega$] with the order topology If $S$ is closed $X \setminus S$ is open in $X$.Then X$\setminus$ S =($\Omega$ )But $\Omega$ has finite countable infinite sets and infinite uncountable set and is open in the order topology and each set is uncountably infinite. Since X\S is uncountably infinite by 5.1 is not bounded above, and since $\Omega$ is a limit ordinal, there is no x$\in$ S for a positive r>0 such that x$\in$(x-r,x+r) Thus Sis not closed in X

I hope I don’t get downvoted. I put a lot of thought into it. I know ordinals are Hausdorff,Conover mentions it, but can’t use it,cause l have to prove it later. Any help would be appreciated

Answer 1

The closedness of $[0,\Omega)$ is equivalent to $\{\Omega\}$ being open. By definition of the order topology, this is equivalent to the fact that $\{\Omega\}$ is an open intervall. (Reason for that: open intervals are a base of the order topology, so $\{\Omega\}$ is open, iff its the union of open intervalls -- since it is a singelton this is equvialent to be exactly one open intervall).

This fails because $\Omega$ has no direct predecessor since it is a limit ordinal.

Answer 2

Let (X,T) be a topology.

This is not a great opening line; in this proof you need $(X,T)$ to be a particular topological space, namely $[0,\Omega]$ with the order topology. It would be better to start with "let $T$ be the order topology on $X = [0,\Omega]$", or something like that, but probably best to just remove this line entirely -- note that you never use $T$ in the rest of the proof, so why introduce it?

If S is closed X\S is open in X

This is a good start to a proof by contradiction! Whenever you start a proof by contradiction, though, you should let the reader know what's happening. For example: "Suppose for the sake of contradiction that $S$ is closed, so that $X \setminus S$ is open" would be more clear.

Pick a point x $\in$ X\S then X\S does not have the property of Theorem 5.1. This means that it is bounded below ,but {0 }is contained in X.

This is confusing for a few reasons:

1. What property in particular are you talking about? Being bounded above? Being countable? It's unclear, so you need to specify.
2. Theorem 5.1 talks about subsets of $[0,\Omega)$. $X \setminus S = \{\Omega\}$ is not a subset of $[0,\Omega)$, so you cannot apply Theorem 5.1 to it. As a side note, you've misquoted the theorem statement in your post: every instance of $[0,\Omega]$ in that statement should be replaced with $[0,\Omega)$.
3. Why did you pick a point $x \in X \setminus S$? You never mention this point again -- if it's necessary for some step in the argument, you need to make that explicit. In any case, the only point in $X \setminus S$ is $\Omega$, so it's weird to make this "arbitrary" choice.
4. Theorem 5.1 is not a biconditional, and only talks about boundedness above. You can use it to show that every subset of $X \setminus S$ which is not bounded above is uncountable, or you can use it to show that every subset of $X \setminus S$ which is countable is bounded above. You certainly cannot use it to deduce that a set which is not a subset of $X \setminus S$ is bounded below (notably, there are sets which are neither bounded above nor bounded below).
5. It's unclear what the fact $\{0\} \subseteq X$ has to do with anything.

This means X\S is not open.

Why? You will need to apply the definition of "open" at some point to deduce this, and it's unclear how you've done so.

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Overall, your proof is not correct. It's unclear what steps you're taking, let alone the justifications for them. I think you should focus on how to prove that a set is not open. You will need to show that $X \setminus S$ does not satisfy the definition of "open set" -- what does this mean?

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Finally, I must point out that the definition of "open set" for the order topology in this book has a technical flaw. Here's the definition in the book:

Let $(X,\leq)$ be a totally ordered set. Then $U \subseteq X$ is open in the order topology on $(X,\leq)$ if and only if for each point $x \in U$, one of the following holds:

1. $x$ is the first point of $X$ and there exists a point $b \in X$ such that the interval $[x,b) = \{p \in X: x \leq p < b\}$ is contained in $U$, or
2. $x$ is the last point of $X$ and there exists a point $a \in X$ such that the interval $(a, x] = \{p ∈ X:a < p \leq x\}$ is contained in $U$, or
3. $x$ is neither the first nor the last point of $X$ and there exist points $a$ and $b$ in $X$ such that $x \in (a, b) = \{p \in X : a < p < b\}$, and $(a, b) \subseteq U$.

The problem with this definition is that the singleton set of a first or last point of a totally ordered set then becomes open:

Proof. First, let $(X,\leq)$ be a totally ordered set with a first point $f \in X$. I claim that $\{f\}$ is open in the order topology. To see this, let $x \in \{f\}$ be arbitrary. Then $x = f$ is the first point of $(X,\leq)$, so we need to show that there exists some $b \in X$ such that $[x,b) \subseteq \{f\}$. We simply pick $b = x$ to get $[x,b) = [f,f) = \varnothing \subseteq \{f\}$. Since $x$ was arbitrary, we conclude that $\{f\}$ is open.

Similarly, if $\ell \in X$ is a last point of $(X, \leq)$, then $\{\ell\}$ will be open, since $(\ell,\ell] = \varnothing \subseteq \{\ell\}$. $\square$

But the whole point of this exercise is that the last point of a totally ordered set need not be open in the order topology! The definition should instead read:

Let $(X,\leq)$ be a totally ordered set. Then $U \subseteq X$ is open in the order topology on $(X,\leq)$ if and only if for each point $x \in U$, one of the following holds:

1. $x$ is the first point of $X$ and there exists a point $b \in X$ such that the interval $[x,b) = \{p \in X: x \leq p < b\}$ is nonempty and contained in $U$, or
2. $x$ is the last point of $X$ and there exists a point $a \in X$ such that the interval $(a, x] = \{p ∈ X:a < p \leq x\}$ is nonempty and contained in $U$, or
3. $x$ is neither the first nor the last point of $X$ and there exist points $a$ and $b$ in $X$ such that $x \in (a, b) = \{p \in X : a < p < b\}$, and $(a, b) \subseteq U$.

Equivalently, we could require that $a,b \neq x$ in the first two cases. It seems like the author accidentally assumed that $[x,b)$ and $(a,x]$ will always include $x$, but this is just not the case. So, please use this corrected definition of "open".