the * property

In two separate meetings, we made judicious use of what Sidney Morris, in his textbook Topology without Tears, calls the * (pronounced ‘star’) property. This property focuses on subsets of R (the set of real numbers), and we can use it to prove many important claims about the topological structure of R and subsets of R.

A subset $S \subset \mathbb{R}$ is said to be open in the euclidean topology on $\mathbb{R}$ if it has the following property: (*) For each $x \in S$ , there exist $a, b \in \mathbb{R}$ with $a < b$ such that $s \in (a,b) \subseteq S$ .

Some practice is recommended to get the hang of this definition, so let’s jump right in.

euclidean toplogy

We have * and want to show that $(\mathbb{R}, \tau)$ satisfies the axioms for a toplogogical space. As a reminder, these axioms are (restated with direct reference to $\mathbb{R}$ without any loss of generality):

the complete set $\mathbb{R}$ and the empty set $\emptyset$ are members of $\tau$
any union of members of $\tau$ is also a member of $\tau$
any finite intersection of members of $\tau$ is also a member of $\tau$

Recall, also, that $\tau$ is simply a collection of subsets of $\mathbb{R}$ .

Proof that 1 holds true for $(\mathbb{R},\tau)$

Pick any real number x. Subtract 1 from x to get a. Add 1 to x to get b. Convince yourself that x lies somewhere ‘between’ a and b.

$$ x \in \mathbb{R}\\ a = x - 1\\ b = x + 1\\ x \in (a, b) $$

Now, $(a, b)$ is an open interval and a subset of $\mathbb{R}$ . This satisfies the definition of * precisely, namely,

$$ x \in (a, b) \subseteq \mathbb{R} $$

and so, by definition, $(a, b)$ is open in $(\mathbb{R}, \tau)$ . Do this for all $x \in \mathbb{R}$ and we show that $\mathbb{R} \in \tau$ .

You can say that for some $y \notin \mathbb{R}$ , there is precisely one interval $\emptyset$ where the * property holds; therefore $\emptyset \in \mathbb{R}$ as well.

Proof that 2 holds true for $(\mathbb{R},\tau)$

Recall that $\tau$ is simply a collection of sets, so we can easily choose some subcollection, call it A. In fact, we can make many such choices; so, each time we choose a unique subcollection of $\tau$ , we can ‘tag’ it (‘index’ is the correct term for this), so we have $A_i$ . All this means is that we can have a collection of subcollections of $\tau$ , and we can keep track of which subcollection we’re talking about using the indices.

To prove the second axiom for $(\mathbb{R},T)$ , we can use the * property again on this family of indexed subcollections $\{ A_i \}$ .

Pick any real number x from a set built from the union of subsets in $A_k$ , which is a specific member of the family $\{ A_i \}$ . We then have x as a member of the subcollection A_k. Now we chose each A such that $A \in T$ . There must then be some $a,b \in \mathbb{R}$ where the statement

$$ x \in (a,b) \subseteq A_k $$

holds true. By induction, this must hold true in general for the family { A_i }, meaning

$$ x \in (a,b) \subseteq A_i $$

holds true. This again is the * property, and so we’ve shown all unions of any subcollection of $\tau$ belongs to $\tau$ .

Proof that 3 holds true for $(\mathbb{R},\tau)$

At this point, we have a bit of freedom – believe it or not, things are getting easier. Pick any pair of subsets $A, B \in \tau$ . Pick some y from the intersection of A and B: $y \in A \cap B$ . By definition of set intersection, $y \in A$ . Using a similar trick as above in proving the second axiom holds on $(\mathbb{R},\tau)$ , we can show

$$ y \in (a,b) \subseteq A

Moreover, y ∈ B by definition of set intersection, and so

$$ y \in (c,d) \subseteq B

Almost there! If we have some number e larger than both a and c, and we have some number f smaller than both b and d; then $e < y < f$ . Hence,

$$ y \in (e,f) \subseteq A \cap B

The intersection – which is, after all, any pair of subset from $\tau$ – has the * property.

extreme generality

The extreme generality involved in this proof caught me off gaurd the first time. But if you think about it, we need an extremely general approach to making a claim about the topological structure of the real numbers – \mathbb{R} is uncountably large! That’s why the proof is so careful to pick, with as little specificity as possible, “any” real number, or “any” set from a subcollection, or “any” pair of subsets to test whether or not it has the * property. If we are able to make this selection with as few restrictions as possible, then the property demonstrated on that selection “carries through” to any other selection we could have made – which is to say any and all open sets.

Topology is like that; its extreme generality allows you to make very careful determinations on the nature of the entire space at once.

Demostrating the ‘Openness’ of Subsets of $\mathbb{R}$

This follows proposition 2.2.1 from Topology without Tears.

A subset $S \subset \mathbb{R}$ is open if and only if it is a union of open intervals.

This claim goes in two directions (“if and only if”); so its strength will be very useful.

Demonstrate `S` is an open set

Assume for the proof that S is a union of open intervals. We construct it with the help of indices i, giving us

$$ S = \cap(a_i, b_i) $$

Luckily, we’ve already proven this above. This immediately gives us S as an open set.

Demonstrate `S` is a union of open intervals

Assume for the proof that S is an open set. Pick some x from S. We’ll use the same trick as above to select an interval (a,b), so

$$ x \in (a,b) \subseteq S $$

Be careful! The temptation here is to say, “use the * property and we’re done”. Yes, we have shown the * property in this case but only in this case! There has been a subtle shift in language that ( if you’re really paying incredibly close attention ) should tip us off that we’ve restricted our selection of x. We’ve chosen “some” not “any” x, and thus we have to justify extending the * property generally across the set S.

The two following step will justify the extension for us.

Demonstrate all `x` is contained in some interval

Pick any x from S. Then we get $x \in (a,b) \subseteq S$ , making us happy. This maneuver is becoming more natural for us.

Demonstrate all intervals contains some `x`

Pick any x from a union of all open intervals. Then x belongs to some open interval (at least one). Then x belongs to S.

`S` is an open set

Why these extra steps? Another subtlty in language: $S = \cup(a_i,b_i)$ is a strong claim; before we may only have claimed that the relation ⊆ rather than = holds for S. ( Recall that the precise meaning of ⊆ is that the set on the left of the symbol is either equal to or a subset of the set on the right of the symbol. This is analogous to the arithmetic relation ≤ which is true when either the two numbers are equal or the number on the left is smaller. )

We demonstrate the * property on some number x of S, generalize it for S, and then generalize it for $\cup(a_i,b_i)$ , giving us justification for = as opposed to the weaker $\subseteq$ ( which we could have satisfied much sooner).

What’s next?

From here, we can move pretty quickly into the definition of a topological basis, a neighborhood for an element in a topological space, and many concrete examples for bases on the euclidiean topology. But let’s let the approach of these proofs soak in first!