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:

1. $a\in M$
2. For all non-decreasing sequences $(x_n{)}_n$ with a limit $x:=lim{x}_n$ (in $\mathbb{R}$), we have $x\in M$
3. 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)
  • $\dots$

These iterations seem to hit the nail, but only informally so far. There is a bit of uncertainty whether $½+\delta +{\delta }^{\prime }+\dots$ 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-closed if 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 $$[m,m_1]⊑[m,m_2]:\iff m_1\le m_2.$$

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

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 $m^{\prime }:=sup{m}_i$. If in fact $m^{\prime }\in T$ is an attained maximum, $[m,m^{\prime }]$ is an upper bound of the chain. Otherwise, $[m,m^{\prime })\in M$ and by upwards-closedness $[m,m^{\prime }]\in M$ is an upper bound.

And finally, we can elegantly¹ 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.