# Short Exercise: Transfinite Closure in Analysis

Recently in the `##math`

chat on Freenode, somebody asked for some intuition and help on the following analysis problem:

Let $a < b$ be reals and $M \subseteq [a, b]$ a subset with these properties:

- $a \in M$
- For all non-decreasing sequences $(x_n)_n$ with a limit $x := \lim x_n$ (in $\mathbb{R}$), we have $x \in M$
- For every $y \in M$ there is a $\delta > 0$ such that $(y - \delta, y + \delta) \cap [a, b] \subseteq [a, b]$
Prove that $M = [a, b]$.

It was quite a **nice exercise for keeping my skills sharp**: the claim is intuitively clear and can be proven more beautifully and less technically than one might think on first sight!

*Update:* I stumbled upon the post *A principle of mathematical induction for partially ordered sets with infima?* on MathOverflow, which presents a generalized induction principle very close to the instantiation I use below.

## The Intuition

Property (2) is quite similar to stating that $M$ is closed in the subspace topology on $[a, b]$ except that it’s only

*upwards-closed*.Property (3) simply says that $M$ is open in $[a, b]$.

Let’s try a few sample $M$s:

- $M = [0, ½)$: contradicts (2)
- fixing that, $M = [0, ½]$: contradicts (3)
- fixing that, $M = [0, ½ + \delta)$: contradicts (2)
- fixing that, $M = [0, ½ + \delta]$: contradicts (3)
- $\ldots$

These iterations seem to hit the nail, but only informally so far. There is a bit of uncertainty whether $½ + \delta + \delta’ + \ldots$ really approaches $b$ after some transfinite amount of steps. In other words, we could see **$[a, b]$ as the transfinite closure of $\varnothing$** wrt. properties (1)-(3).

Morally, these iterations always take **the longest strip** so far and make that even longer: if the longest strip so far is $[a, c]$, then by (3) make it into $[a, c + \delta)$ and in turn by (2) into $[a, c + \delta]$. **That reminded me heavily of a partial order and of Zorn’s lemma**! Indeed, it turns out that this leads to a correct formal proof as elaborated below.

## The Formalism

To make the proof smoother and less notation-heavy, let us rephrase property (2) a bit:

Definition:We call a subspace $M \subseteq \mathbb{R}$upwards-closedif all non-decreasing Cauchy sequences in $M$ have a limit in $M$.

Now we can state the poset of strips and prove the existence of a maximal element by Zorn’s lemma:

Definition (Poset of Strips):For a pointed subset $(M, m) \subseteq \mathbb{R}$ define a partial order on the set of strips $[m, \cdot]$ with $\cdot \in M$ such that $$[m, m_1] \sqsubseteq [m, m_2] :\Leftrightarrow m_1 \leq m_2.$$

Lemma:Upwards-closed pointed subsets $(M, m)$ have a longest strip $[m, m’] \in M$.

Proof:Apply Zorn’s lemma. Let $T = (m_i)_{i \in I}$ be a chain in the poset of strips. If $T$ is empty, then $[m, m]$ is a trivial upper bound. Otherwise, define $m’ := \sup m_i$. If in fact $m’ \in T$ is an attained maximum, $[m, m’]$ is an upper bound of the chain. Otherwise, $[m, m’) \in M$ and by upwards-closedness $[m, m’] \in M$ is an upper bound.

And finally, we can elegantly^{1} prove the overall claim:

Theorem:Let $a < b$ be two real numbers and $(M, a)$ a subset $M \subseteq [a, b]$ that is open and upwards-closed therein. Then $M = [a, b]$.

Proof:The previous lemma guarantees a longest strip $[a, c] \subseteq M$. We show $c = b$ by assuming the contrary and deriving a contradiction. Since $M$ is open, there is $\delta > 0$ such that $[a, c + \delta) \subseteq M$. Due to upwards-closedness, we conclude that $[a, c + \delta] \subseteq M$ is an even longer strip contradicting the maximality given by the lemma.

- Well, for some people it might be debatable how elegant (constructive) an invocation of Zorn’s lemma is.
^{^}