Give a "constructive" proof of the fact that in a metric space the intersection of two open balls is open

Question

Main Question

Can someone give a "constructive" proof of the fact that,

Let $(X,d)$ be a metric space and $x,y\in X$. Let $B_d(x,r_x)$ and $B_d(y,r_y)$ be two open balls centered respectively at $x$ and $y$ for $r_x,r_y\in \mathbb{R}^+$. Suppose that $B_d(x,r_x)\cap B_d(y,r_y)\ne \emptyset$ and $z\in B_d(x,r_x)\cap B_d(y,r_y)$. Show that there exists $r_z\in\mathbb{R}^+$ such that $$B_d(z,r_z)\subseteq B_d(x,r_x)\cap B_d(y,r_y)$$

I have actually tried to give a constructive proof of the fact in this post (see the "Second Proof") but after reading the proof carefully I came to the conclusion that it doesn't satisfy my aim because it the second proof doesn't actually "constructs" something, it actually gives a motivation for it. I am not sure whether it is the same thing as a "constructive" argument.

Relevant Definitions

Definition of Open Ball Let $(X, d)$ be a metric space and let $r\in\mathbb{R}^+$. Then the set, $B_d(x, r) := \{y \in X : d(x, y) < r\}$ will be said to be the open ball of radius $r$ centered at $x$ in the metric space $(X, d)$.

Definition of Open Set

Let $(X,d)$ be a metric space and $U\subseteq X$. Then $U$ will be said to be $d$-open in $X$ if for all $x\in U$, there exists $r>0$ such that $B_d(x,r)\subseteq U$.

It may be asked that what I have meant by a "constructive" proof. For that purpose, I think that the following will illustrate what I mean by a "constructive" proof.

Proposition Let $(X, d)$ be a metric space, $x \in X$ and $r > 0$. Then the open ball $B_d(x,r)$ is open.

Proof Assume an $s>0$ exists such that $B_d(y,s)\subseteq B_d(x,r)$. If $z \in B(y, s)$, we want to show that $d(z, x) < r$. Now, $$d(z, x) \le d(z, y) + d(y, x) <s + d(x, y) < r$$ This prompts us to consider $0 < s <r - d(x, y)$. Let $s$ be one such. If $z \in B(y,s)$, then we have $$d(z, x) < d(z, y) + d(y, x) <s + d(y, x) < r$$ Thus, $B_d(y, s) \subseteq B_d(x,r)$. Since $y$ is an arbitrary element of $B_d(x , r)$, we are done.

Answer 1

You need to be aware that whatever you mean by "constructive" is very different from what the word usually means. (You "need" to know this, because when you use words meaning something other than what they usually mean that's going to lead to confusion. Confusion when people read what you write and when others read what you write.)

So for the record, in this post I'm using the word constructive with its usual meaning - if I'm referring to what you mean, which nobody quite follows, I'l use scare quotes: "contructive".

In particular there is absolutely nothing non-constructive about the standard proof that the intersection of two open balls is open. Instead of telling Brian that the proof is not constructive, you'd be better off trying to understand how it is constructive; this might help you clarify for yourself the difference between constructive and "constructive".

Consider the following two locutions:

1. The set $A$ is non-empty.

2. Let $x\in A$; then [blah blah blah whatever].

As far as I can tell from your comments, you seem to feel that going from (1) to (2) is somehow not "constructive". But people who talk about constructive this and that do not object to arguing as in (2), if (1) is given: (1) means that there exists $x\in A$, and nobody but you sees any harm in saying "ok, since we're given that there exists $x\in A$, let's assume that $x\in A$".

Where actual constructivists differ from most mathematicians is in how, for example, one might establish (1) in the first place; for example most people would say that deriving a contradiction from $A=\emptyset$ counts as a proof of (1), while a constructivist would say no, that's not enough, to show that $A$ is nonempty you have to "construct" or otherwise explicitly exhibit an element of $A$. But nobody but you finds anything non"constructive" in passing from (1) to (2), once (1) has been established, or if (1) is given.

Similarly with these open balls. Consider

3. $x\in B(y,r)$.

4. $\delta=r-d(x,y)>0$.

Your objection to deriving (4) from (3) seems incoherent to the rest of us; if the objection has something to do with something being not "constructive" fine, we don't know what that means so we can't say But it's a fact that deriving (4) from (3) is perfectly acceptable constructively. Because $d(x,y)<r$ is simply what (3) means! If we're given (3) then your idea that we need to "construct" $d(x,y)$ seems simply incoherent.

Answer 2

Hint: in constructive reasoning the data (i.e., the objects appearing in the assumptions of your theorem) are given constructively. A real number for example, would be given as a computable real. A metric space $(X, d)$ would be given as a pair of computable functions: one that tests for membership of $X$ and one that calculates $d(x, y)$ for $x, y \in X$. An open set $A$ would be given as a pair of computable functions: one that tests for membership of $A$ and one that given $x \in A$ returns a computable real $r > 0$ such that $B(x, r) \subseteq A$. Hence the standard proofs of the following facts

• an open ball is an open set
• the intersection of two open sets is an open set

are constructive, since the values you need as witnesses in the proof are computable functions of the input data.

An example of a non-constructive proof is the existence of the supremum of a non-empty bounded above set $X$. The data would comprise the membership test for $X$, an element of $X$ and a bound for $X$. The existence of Specker sequnces shows that there no way to compute the supremum from this data.

Answer 3

What you give as an illustration of what you mean by a "constructive proof" actually illustrates that you're not talking about proofs at all; my conjecture is that you're really talking about an explanation of how one might find a proof of something.

Your illustration:

Proposition Let $(X, d)$ be a metric space, $x \in X$ and $r > 0$. Then the open ball $B_d(x,r)$ is open.

Proof Assume an $s>0$ exists such that $B_d(y,s)\subseteq B_d(x,r)$. If $z \in B(y, s)$, we want to show that $d(z, x) < r$. Now, $$d(z, x) \le d(z, y) + d(y, x) <s + d(x, y) < r$$ This prompts us to consider $0 < s <r - d(x, y)$. Let $s$ be one such. If $z \in B(y,s)$, then we have $$d(z, x) < d(z, y) + d(y, x) <s + d(y, x) < r$$ Thus, $B_d(y, s) \subseteq B_d(x,r)$. Since $y$ is an arbitrary element of $B_d(x , r)$, we are done.

This is simply not in any way shape or form a proof that open balls are open. To prove that you need to show that if $z\in B(x,r)$ then there exists $s>0$ such that $B(z,s)\subset B(x,r)$. And you simply do not prove that. A proof of that cannot begin by assuming that $B(z,s)\subset B(x,r)$.

One might regard it as an illustration of how one might find a proof of the proposition. If so it's very badlyy phrased - over and over you say $A$ implies $B$ when what you mean is more something like "We need to prove $A$; now that would follow from $B$" or something.

Assuming that it's actually supposed to be motivation for the proof, one might give a much less incoherent presentation:

Motivation: We need to show that if $z\in B(x,r)$ then there exists $s>0$ so that $b(z,s)\subset B(x,r)$. So we assumme that $z\in B(x,r)$. Now how could we find an $s>0$ as required? We need to show that if $w\in B(z,s)$ then $w\in B(x,r)$, which is to say we need to show there exists $s>0$ with this property: If $d(w,z)<s$ then $d(x,w)<r$. But if we know that $w\in B(z,s)$ then the triangle inequality will show that $$d(x,w)\le d(x,z)+d(z,w)<d(x,z)+s;$$hence we only need to show that there exists $s>0$ such that $d(x,z)+s<r$. Since $d(x,z)<r$ it is clear that $s=r-d(x,z)$ will suffice.

Or something like that - I'm not going to bother proofreading the above. The point is just the difference between starting a supposed "proof" by assuming exactly what is to be proven, and starting an explanation of how one might find a proof by stating clearly what is to be proven, then commenting on what might prove that.