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Rudin's proof that 'every neighbourhood is open' goes as follows:

By the supposition, for some metric space $X$ with metric d, there exists a neighborhood $N_r$ around $p$ with some radius $r$.

For all points $q \in N_r,\;d(p,q)=r-\varepsilon<r$ for some positve $\varepsilon$. Consider a point $s$ in $X$ such that $d(s,q)<\varepsilon$. Then:

$$d(p,s)\leq d(p,q)+d(q,s)<r-\varepsilon\;+\varepsilon$$ $$\implies d(p,s)<r$$

Such an argument applies for all points in the neighbourhood $N_\varepsilon$ around point $q$. So all points in $N_r$ are interior points and thus $N_r$ is open.

My question originates on whether we know for certain that such a point $s$ exists. Consider for example the subset of $\mathbb{R}$:

$$X=\{0, 0.2, 0.9\}$$

With $p=0$, $q=0.9$, $r=1$ and $\varepsilon=0.1$. There is no point $s$ in the neighborhood such that $d(q,s)<\varepsilon$.

My question: How does the proof hold if the proposed neighborhood $N_\varepsilon$ around $q$ can be empty (note by Rudin's definition, a neighborhood around a point doesn't include the point itself)?

Definitions:

  • A neighborhood with radius $r$, $N_r$ around $p$, is the set: $\,N_r=\{q\;|\;d(p,q)<r\}$ for some $r>0$.

  • An interior point of set E is one such that a neighborhood exists around it that is a subset of E.

  • An open point is a set such that all points within it are interior points.

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  • $\begingroup$ Well, there's at least one such point $s$: just take $s$ to be the point $q$ itself (we have $0 = d(q,q) < \varepsilon$). But I don't think this really answers your underlying question...? $\endgroup$
    – The Amplitwist
    Commented 2 days ago
  • $\begingroup$ @TheAmplitwist By Rudin's definition, {q} isn't a neighborhood around q as r must be greater than zero. $\endgroup$
    – iambadatnames
    Commented 2 days ago
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    $\begingroup$ You wanted to know whether any point $s$ exists in $N_\varepsilon$, which is a neighborhood of the point $q$. My comment says that $q$ belongs to the neighborhood $N_\varepsilon$. I do agree that $N_\varepsilon \neq \{ q \}$ for any $\varepsilon > 0$, here. Does this help? $\endgroup$
    – The Amplitwist
    Commented 2 days ago
  • $\begingroup$ My question was how the above proof is valid considering $N_\varepsilon$ is empty. $\endgroup$
    – iambadatnames
    Commented yesterday
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    $\begingroup$ Nowhere in his proof did he ever claim $s\ne q$ nor $s\ne p$ nor $p\ne q$. In your case, the open neighorhood $N_{q, r} = \{q\}$ and to find an $s$ so that $d(s,q) < 0.1$ simply let $s = q$. Then $d(s,q)=d(q,q)=0 < \epsilon$ and the proof works just fine. $\endgroup$
    – fleablood
    Commented yesterday

3 Answers 3

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Rudin wants to show that if $N_r$ is given to be a neighborhood of some point $p \in X$, then every point $q \in N_r$ has a neighborhood $N_\varepsilon$ contained in $N_r$ (where $\varepsilon$ depends on $q$ here). Rudin takes as a candidate for $\varepsilon$ the number $r - d(p,q)$, which is always positive. Now, how does one show that $N_\varepsilon$ is contained in $N_r$? You pick an arbitrary member of $N_\varepsilon$, say $s$, and show that $s$ is also a member of $N_r$. If you succeed, then since $s$ was arbitrary you can conclude that $N_\varepsilon$ is a subset of $N_r$. This is what is going on in the second half of the proof.

Is it possible that no such $s$ exists that belongs to $N_\varepsilon$? No. As per Rudin's definition (as you have quoted it in the question), any neighborhood $N_r$ of a point $p$ contains the point $p$ itself, since $d(p,p) = 0$ and $0 < r$. So, there is always at least one such point.

On the other hand, even if no such $s$ existed, there is no problem. It just means that the property you want to prove is vacuously true.

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Proving a "for all/every" statement (e.g., openness of a set) never requires existence of any elements of the given set.

As indicated at the end of another answer, the truthfulness of a "for all" statement in regards to an empty set in this manner is called a vacuous truth. I think this is really the crucial conceptual gap that the OP's question suggests. The matter of whether the set is empty or not is simply a nonissue.

Consider the empty set. Every possible "for all" statement about this set is automatically true. For example, let's say you have the predicate "is in a neighborhood around $p$". Is it the case that every element in the empty set is in the neighborhood? Yes, every element I ever pick (namely, nothing) is in the neighborhood.

It might help to think about the converse: To prove a for-all statement false, you need to show a counterexample, that is, that there exists an element for which the predicate is false. Does that happen in this example? No, since the set is empty, no such counterexample exists. Hence the for-all statement is true (vacuously).

In the particular case of the OP's question here, they've also misinterpreted the fact that center points are included in neighborhoods, so the sets can't be empty — but from my perspective that's a secondary issue, because that part doesn't matter in any case.

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A neighborhood can't be empty as it always contains the center point.

Rudins proof is just fine.

Consider $X= \{0, 0.2, 0.9\}$. And $p=0$ and we want to prove $N(0)_{0.1}$ is an open set. ($p = 0$ and $r = 0.1$)

What is $N(0)_{0.1}$? $N(0)_{0.1} = \{q\in X| d(q,0) < 0.1\} = \{0\}$. We have to prove $\{0\}$ is open.

Suppose $q \in \{0\}$. (Well, that means $q = 0$.) Let $d(q,0) = 0.1 -\epsilon < 0.1$. That is to say, $\epsilon = 0.1 - d(q,0) = 0.1-d(0,0) = 0.1 - 0 = 0.1$. So $\epsilon = 0.1$.

Let $s \in N(q)_{\epsilon} = \{x\in X| d(q,x) < \epsilon\} = \{x\in X| d(0, x) < 0.1\} = \{0\}$. So $s = 0$.

Now we need to prove that $s \in N(p)_r = N(0)_{0.1} = 0$. (We already know that because $s = 0$!) but lets do rudin's proof.

$d(p,s)\le$ (remember $p=0;s=0$)

$d(p,q) + d(q,s)$ (remember $q=0$)

$< (r-\epsilon) + \epsilon$ (remember that $\epsilon=r- d(p,q)= r-0=r$ and $d(s,p)=0 < \epsilon$)

$= r$.

So $N(0)_{0.1} = \{0\}$ is an open set.

EVERYTHING works just fine. Just... trivially so.

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I think I misunderstood your counterexample.

I think you may have been saying somithing like this.

Let $X = \{0.2, 0.9, 1\}$ with $p = 1$ and $r= 0.15$.

Then $N(p)_{r}=N(1)_{0.15}= \{0.9, 1\}$. We need to prove $\{0.9, 1\}$ is open.

Rudin says let $q\in \{0.9, 1\}$. We have two cases: $q=0.9$ or $q=1$.

Case A: $q=0.9$. Rudin says let $d(p,q)=r-\epsilon< r$. That is to say $d(p,q) = 0.1$ and $r=0.15$ for $\epsilon = 0.05$.

Case B: $q=1$ and so $d(p,q)=d(0,0) =0$ so $\epsilon = r= 0.15$

Rudin says to consider $s$ so that $d(q,s) < \epsilon$.

Case A: $q=0.9$ and $\epsilon = 0.05$. So we need $d(s,0.9) < 0.05$ so $0.85 < s < 0.95$ so $s = 0.9$ is the only option.

Case B: $q = 1$ and $\epsilon = 0.15$ We need $d(s,1) < 0.15$. So $0.85 < s < 1.15$ there are two choices. B: $s = 0.9$ or C: $s = 1$.

$d(s,p)= d(s,1) < r= 0.15$.

Case A: $q=0.9; \epsilon = 0.05; s=0.9$. Then $d(s,p) = d(0.9, 1) = 0.1 < 0.15.

(If we wanted to be pedantic and use the triangle inequality, then $d(s,p) < d(s,q) +d(q,p)=0+ 0.1 < \epsilon + (r-\epsilon) = 0.05 + (0.15-0.05) = 0.15$.)

Case B: $q = 1; \epsilon = 0.15; s=0.9$. Then $d(s,p) = d(0.9,1) = 0.1 < 0.15$.

Or with triangle inequality $d(s,p) < d(s,q) + d(p,q)= 0.1 + 0 < \epsilon +(r-\epsilon) = 0.15+(0.15-0.15) = 0.15$.

Case C: $q =1; \epsilon =0.15; s=1$. The $d(s,p)=d(1,1)=0 < 0.15$.

Or $d(s,p) < d(s,q) + d(p,q) = 0 + 0 < \epsilon + (r - \epsilon) = 0.15 + (0.15-0.15)= 0.15$

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