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I would like to prove that an open ball is an open set. I have seen the proof here, but I would like to know exactly why is my proof incorrect.

The definition of an open set in my textbook is the following:

A subset $U$ of a metric space $(X,d)$ is open (in the vector space $X$) if for every $x\in U$, there exists $\epsilon>0$ such that $B(x,\epsilon)\subset U$.

Here's my proof:

Let $B(a,r)=\{x\in X:d(x,a)<r\}$ be an open ball in $X$.

Then for every $x\in B$ there exists $\epsilon \in (0,r]$ such that $B(x,\epsilon)\subset B(a,r)$.

For me this seems to satisfy the definition. But it's wrong probably. Why?

  • It is true, but you have not proved it. –  Jun 22 '21 at 10:28
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    You have to explain why such $\epsilon$ indeed exists. This is easy, but not absolutely obvious. – Mark Jun 22 '21 at 10:28
  • @Mark i'm not sure why can't we just take $\epsilon \in (0,r]$? Since if $\epsilon$ is in that interval, it satisfies every $x$ to be in a ball $B(x,\epsilon)$ within $B(a,r)$ What else is there to explain? I don't completely understand. Sorry i'm new to such proofs/explanations –  Jun 22 '21 at 10:35
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    Draw a picture for the two dimensional case. It may help you see how to do the general case. – coffeemath Jun 22 '21 at 10:35
  • $X$ is a vector space and a metric space? – Adam Rubinson Jun 22 '21 at 10:39
  • @AdamRubinson $(X,d)$ is a metric space. ($X$ equipped with the metric $d$) –  Jun 22 '21 at 10:39
  • @ДарияСоколова If your assertion is that for all metric spaces $(X,d)$, for all $a\in X$, for all $r>0$, for all $x\in B(a,r)$ and for all $0<\epsilon\le r$, $B(x,\epsilon)\subseteq B(a,r)$ (which is different from what you've written) then you are wrong. See $X={1,3,5}$, $d(t,s)=\lvert t-s\rvert$, $a=1$, $r=4$, $x=3$, $\epsilon=3$. –  Jun 22 '21 at 10:42
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    You can't take any $\epsilon\in (0,r]$. This will work for $x=a$, but not for every $x$. For example: take the interval $(-1,1)$ in $\mathbb{R}$, which is the ball $B(0,1)$ in that space. Note that the point $x=\frac{1}{2}$ belongs to that ball, but $B(\frac{1}{2},1)$ is not contained in $B(0,1)$. So for the point $x=\frac{1}{2}$ you will need a smaller radius. – Mark Jun 22 '21 at 10:42
  • @Mark okay, I see why now –  Jun 22 '21 at 10:45
  • Then You have written “Vector Space” by accident half-way through your question and may want to edit this to say “metric space”. – Adam Rubinson Jun 22 '21 at 10:48
  • @AdamRubinson actually it's written exactly like that in my textbook. It would be the same thing essentially, no? Maybe they just forgot (typo), so please explain your reasoning why you think it's not correct to write it like that, I am interested to know. –  Jun 22 '21 at 10:52
  • For the difference between metric spaces and vector spaces, see: https://math.stackexchange.com/questions/114940/what-is-the-difference-between-metric-spaces-and-vector-spaces – Adam Rubinson Jun 22 '21 at 11:25

2 Answers2

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Take the open ball $B(a,r)=\{x\in X: d(x,a)<r\}$. For $y\in B(a,r)$ let $r_1=r-d(y,a)$. Then if $x\in B(y,r_1)$ we have

$d(x,a) \le d(x,y)+d(y,a)<r_1+d(y,a)=r$

which implies that the ball $B(y,r_1)$ is contained in $B(a,r)$ and thus the given ball must be open.

  • thank you, but that's the answer to the question I cited. –  Jun 22 '21 at 10:47
  • Whoops, sorry. I hope it is of at least some help though. Try to draw a picture of it in two or three dimensions, that would help a lot. There are some available online if you look for them. –  Jun 22 '21 at 10:53
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Here's my proof:

Let $B(a,r)=\{x\in X:d(x,a)<r\}$ be an open ball in $X$.

Then for every $x\in B$ there exists $\epsilon \in (0,r]$ such that $B(x,\epsilon)\subset B(a,r)$.

For me this seems to satisfy the definition. But it's wrong probably. Why?

A proof is when you provide convincing reasons/ arguments for the proposition. You're asserted that $\varepsilon$ exists, without giving convincing reasons that it does.

Your "proof" is similar to the following (insufficient) "proof" for why there isn't a greatest real number less than $0$:

Suppose $\ x<0\ $ is the greatest real number less than $\ 0.\ $ Then we can find $\ y\ $ such that $\ x<y<0.\ $ This contradicts that maximality of $\ x,\ $ and so we have a contradiction. Therefore no such "greatest real number less than $\ 0"\ $ exists.

This may seem convincing at first, but if you look again, you will see that we have merely asserted that $\ \exists\ y\ $ such that $\ x<y<0.\ $ We haven't actually shown that there exists such a number. So we haven't proven anything. The reason the above "proof" may seem convincing is because we know it works if we fill in the missing part. However, we shouldn't leave out this missing part in our proof. The missing part in the above proof is, for example, $\ y= \frac{1}{3}x.\ $ Then we can write: $\ y-x = \frac{1}{3}x-x = -\frac{2}{3}x > 0\ $ (using the axiom that $\ a.b>0\ $ if $\ a>0\ $ and $\ b>0\ $), therefore $\ y>x.\ $ And $\ y=\frac{2}{3}x<0\ $ because $\ \frac{2}{3}>0\ $ and $\ x<0,\ $ so here we are using the axiom that $\ a.b<0\ $ if $\ a>0\ $ and $\ b<0.\ $ Okay, so now we have actually shown that $\ x<y<0\ $ is always true if we set $\ y=\frac{1}{3},\ $ we can add this into our proof and our proof is fine.

You need to do something similar in your proof that convinces us that you have actually found $\ \varepsilon\ $ such that $\ B(x,\varepsilon)\subset B(a,r).$

Adam Rubinson
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