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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 $B_d(x,r)\subseteq U$.

Problem. Let $(X,d)$ be a metric space and $x,y\in X$. Let $B_d(x,r)$ and $B_d(y,s)$ be two open balls in $X$. Show that $B_d(x,r_x)\cap B_d(y,r_y)$ is open in $X$.

First Proof.

If $B_d(x,r_x)\cap B_d(y,r_y)=\emptyset$ then we have nothing to prove since $\emptyset$ is open in $X$.

Otherwise, $B_d(x,r_x)\cap B_d(y,r_y)\ne\emptyset$ and let $z\in B_d(x,r_x)\cap B_d(y,r_y)$. Choose $r_z=\min\{r_x-d(x,z),r_y-d(y,z)\}$. Now consider the open ball $B_d(z,r_z)$. Let $u\in B_d(z,r_z)$. Then we have, $$d(x,u)\le d(x,z)+d(z,u)<r_x$$which shows that $u\in B_d(x,r_x)$. Also, $$d(y,u)\le d(y,z)+d(z,u)<r_y$$which shows that $u\in B_d(y,r_y)$. Consequently we have, $u\in B_d(x,r_x)\cap B_d(y,r_y)$. So, we conclude that $B_d(z,r_z)\subseteq B_d(x,r_x)\cap B_d(y,r_y)$ and we are done.

Second Proof.

If $B_d(x,r_x)\cap B_d(y,r_y)=\emptyset$ then we have nothing to prove since $\emptyset$ is open in $X$.

Otherwise, $B_d(x,r_x)\cap B_d(y,r_y)\ne\emptyset$ and let $z\in B_d(x,r_x)\cap B_d(y,r_y)$. If we can show that there exists $r_z\in\mathbb{R}^+$ such that $B_d(z,r_z)\subseteq B_d(x,r_x)\cap B_d(y,r_y)$ then we are done.

So, let us assume that such $r_z$ exists and let $u\in B_d(z,r_z)$. Then, $$d(x,u)\le d(x,z)+d(z,u)<d(x,z)+r_z$$If we want that $d(x,z)+r_z\le r_x$ Then we choose $r_z$ to be the infimum of the set, $$R:=\{r_v\in\mathbb{R}:(\exists v\in B_d(x,r_x)\setminus\{x\})[r_v=r_x-d(x,v)]\}$$ provided $R\ne \emptyset$. Such infimum exists since $R$ is bounded below by $0$. Since $r_z=\inf R$ we conclude that $r_z\le r_x-d(x,v)$ for all $v\in B_d(x,r_x)$. Hence in particular for $v=z$ we have $r_z\le r_x-d(x,z)$. We conclude that if such $r_z$ exists then $r_z\le r_x-d(x,z)$.

In a similar manner we will be able to show that $r_z\le r_y-d(y,z)$. Consequently we get, $$r_z\le \min \{r_x-d(x,z),r_y-d(y,z)\}$$i.e., if such an $r_z$ exists then $r_z\le \min \{r_x-d(x,z),r_y-d(y,z)\}$.

Now let us choose $r_z\in (0,\min \{r_x-d(x,z),r_y-d(y,z)\}]$. Now consider the open ball $B_d(z,r_z)$. Let $u\in B_d(z,r_z)$. Then we have, $$d(x,u)\le d(x,z)+d(z,u)<r_x$$which shows that $u\in B_d(x,r_x)$. Also, $$d(y,u)\le d(y,z)+d(z,u)<r_y$$which shows that $u\in B_d(y,r_y)$. Consequently we have, $u\in B_d(x,r_x)\cap B_d(y,r_y)$. So, we conclude that $B_d(z,r_z)\subseteq B_d(x,r_x)\cap B_d(y,r_y)$ and we are done.

There is an important difference in the two proofs. In the first proof we don't need to consider the case when $r_z=0$ but in the second proof we need to consider that (although I haven't done that, and I think that it is a flaw) because $\inf R$ can be $0$. But then we need to define what do we mean by an open ball of radius $0$. Probably that will be $\emptyset$ as has been suggested here. The question is,

Is it "wrong" to define an open ball in a metric space $(X,d)$ as has been defined above? If not then how can one resolve the apparent paradox that results from the second proof?

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    In your second proof, you are assuming that such an $r_{z}$ exists. That is what you are trying to prove so how can you assume it exists? – Jack Feb 06 '16 at 06:03
  • @Tim: It seems that you haven't read the complete proof. First, I assume that such an $r_z$ exists and deduce the range that this $r_z$ must satisfy if it exists. Then I pick up an arbitrary $r_z$ from the range and show that it indeed satisfies the condition that $B_d(z,r_z)\subseteq B_d(x,r_x)\cap B_d(y,r_y)$. This proves the existence of $r_z$. –  Feb 06 '16 at 06:08
  • @Tim: In other words, I am "working backwards". –  Feb 06 '16 at 06:12
  • @Tim: Let us assume that I have $5$ fingers. then I can curl them. We conclude I can catch a baseball.-This is not the type of argument I have used. In your argument you have only proved that, if you have $5$ fingers then you can catch a baseball however it needs to be shown that you indeed have $5$ fingers. Read my previous comment again. –  Feb 06 '16 at 06:31
  • I deleted my comment because i realized it wasn't very good. However, if you want to prove such an $r_{z}$ exists, then you should just construct it, not assume it. – Jack Feb 06 '16 at 06:33
  • @Tim: Sure. The second proof indeed "constructs" it in some sense. –  Feb 06 '16 at 06:34
  • You don't need to consider $r_z$ = 0. $r_z > 0$ by definition. If however if $\inf R$ = 0, then we have proven that $r_z$ can not exist and the statement is false. At this point with haven't defined $r_z$ or proven it exists. Then when we start the proof we select an $r_z$ from an interval that does not include 0. Perhaps we shouldn't have called it $r_z$. Id have called it $r_w$ or something. Then we prove that is satisfies the conditions for the $r_z$. We don't consider $r_z = 0$; we just consider $r_z > 0$ or $r_z$ doesn't exist. – fleablood Feb 06 '16 at 07:44
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    Even if we considered $r_z = 0$ as you interpret it, I don't see what the paradox implied is. – fleablood Feb 06 '16 at 07:46
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    If $R\ne \phi$ then $0\leq \inf R<r_x$ but then, for $R\ne \phi$ we have $\inf R>0\iff B_d(x,r_x)=B_d(x,s)$ for any $s\in (r_x-\inf R,r_x)$, which is possible but not necessarily true. And when trying to find a suitable open ball about $z$,we don't want to consider $R$. We just want to consider points in $B_d(x,r_x)$ that are close to $z$.Also an open ball of radius $0$ is an open ball,but it's also the empty set, so $z$ is not a member of it. I suggest that the second proof is too muddled to waste time on. – DanielWainfleet Feb 06 '16 at 08:12

2 Answers2

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In the second proof, everything between "So, let us assume..." and "...then $r_z\le \min \{r_x-d(x,z),r_y-d(y,z)\}$." is not really part of the proof (you may notice that nothing proven in that section is ever used to make any logical deduction in the remainder of the proof). Rather, it is motivation for the following line of the proof: "Now let us choose $r_z\in (0,\min \{r_x-d(x,z),r_y-d(y,z)\}]$". The preceding section attempts to show why this is a wise choice of $r_z$, but you could still make this choice of $r_z$ even if you didn't provide any reason to think it was a wise choice (which is what is done in the first proof).

In particular, you don't care if $\inf R$ could be $0$, because nothing about $R$ is actually necessary for the validity of the second proof.

[In fact, the motivation provided by the discussion about $R$ is not actually good motivation. In most examples, $\inf R$ actually will be $0$, and so it was wrong to think that $r_z$ was forced to be $\inf R$. Indeed, it is unclear to me why you defined $R$ at all, or why you think $\inf R$ is a good choice for $r_z$.]

Eric Wofsey
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  • I don't understand why "nothing about $R$ is actually necessary for the validity of the second proof" because $\inf R$ can be $0$. –  Feb 06 '16 at 06:19
  • And if $\inf R=0$ then my choice of $r_z$ prompts us to choose $r_z=0$. –  Feb 06 '16 at 06:20
  • But you don't actually make that choice of $r_z$! Instead, you choose $r_z\in(0,\min {r_x-d(x,z),r_y-d(y,z)}]$. – Eric Wofsey Feb 06 '16 at 06:27
  • That's why I said that (read the post again) I think it is a flaw. –  Feb 06 '16 at 06:28
  • Sorry, I don't understand what you meant by your previous comment (I don't know what "that" is referring to, or what I'm supposed to see when rereading the post, or what "it" is referring to). – Eric Wofsey Feb 06 '16 at 06:32
  • I think that the 'paradox' lies in the fact that in the second proof we can actually choose $r_z=0$ (at least I don't see any reason for avoiding this possibility) but in the first proof we can't. –  Feb 06 '16 at 06:35
  • The definition of "open set" (well, a definition; I'm not sure which definition you're using) is that a set $U$ is open iff for each $z\in U$, there exists an $r_z>0$ such that $B_d(z,r_z)\subseteq U$. So if you choose $r_z=0$, you will fail to satisfy the definition, and so the proof fails. (If you are using some different definition, you should state what that definition is.) – Eric Wofsey Feb 06 '16 at 06:38
  • Let us continue this discussion in chat. –  Feb 06 '16 at 06:41
  • see my asnwer why the second proof is flawed. – runaround Feb 06 '16 at 06:48
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The second proof is flawed: The defintion is R is incorrect $$R:=\{r_v\in\mathbb{R}:(\forall v\in B_d(x,r_x))[r_v=r_x-d(x,v)]\}$$. I guess it might be $$R:=\{r_v\in\mathbb{R}:(\exists v\in B_d(x,r_x))[r_v=r_x-d(x,v)]\}$$

Also, the infimum of R could be zero. A ball with radius 0 is not open ball. The mistake is the set R is defined too broad so it is useless in the proof.

runaround
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  • The person who voteds this down is wrong. The second proof is flawed. – runaround Feb 06 '16 at 06:40
  • It is true that the definition of $R$ is stated wrong, but this is totally irrelevant to the validity of the second proof; see my answer. – Eric Wofsey Feb 06 '16 at 06:40
  • We are going find out why the second proof is wrong. My answer is correct, your answer is off topic. – runaround Feb 06 '16 at 06:44
  • I don't understand what is wrong in the definition of $R$, can it be clarified? –  Feb 06 '16 at 06:58
  • @user170039: You are trying to define $R$ to be the set of all values of the form $r_x-d(x,v)$, where $v$ can be any element of $B_d(x,r_x)$. But by your definition, $r\in R$ iff $r=r_x-d(x,v)$ for every value of $v$ simultaneously! If there are two points $v,v'\in B_d(x,r_x)$ such that $d(x,v)\neq d(x,v')$, then there is no such $r$, so $R$ is empty by your definition. To fix this, you need to replace the $\forall$ by $\exists$ in your definition of $R$. – Eric Wofsey Feb 06 '16 at 07:02
  • You use $\forall v\in B_d(x,r_x)$ in defining R. So R could be empty. – runaround Feb 06 '16 at 07:02
  • if you restrict set R to a proper set, it will be back to the first proof. – runaround Feb 06 '16 at 07:05
  • @EricWofsey: I don't understand what you meant when you said "[b]ut by your definition, $r∈R$ iff $r=r_x−d(x,v)$ for every value of $v$ simultaneously!". Also I don't understand why "[i]f there are two points $v,v′∈B_d(x,r_x)$ such that $d(x,v)≠d(x,v′)$, then there is no such $r$". Wouldn't then $r_v,r_v'\in R$? –  Feb 06 '16 at 07:12
  • You may say that I need to consider $B_d(x,r_x)\setminus{x}$ instead of $B_d(x,r_x)$ (which I have edited already) but then the proof becomes similar to what I have already mentioned here. –  Feb 06 '16 at 07:17
  • @user170039: Your definition is that $r_v\in R$ iff $(\forall v\in B_d(x,r_x)) r_v=r_x-d(x,v)$. Replacing $r_v$ with $r$ to avoid notational confusion and stating the "for all..." in words instead of symbols, this is exactly what I said in the the first quote you are asking about. – Eric Wofsey Feb 06 '16 at 07:18
  • @EricWofsey: I see now. Thanks for explaining. –  Feb 06 '16 at 07:23
  • @EricWofsey: Is it ok now? –  Feb 06 '16 at 07:25
  • @user170039: The definition now makes more sense. However, the reasoning you are attempting to make with $R$ is still mysterious to me (see the parenthetical I added to the end of my answer). Nevertheless, since you never actually use the definition $r_z=\inf R$ in the second proof, the proof is still valid. – Eric Wofsey Feb 06 '16 at 07:46