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I'm confused about the definition of Open set "A set Z is open, if for every element z in Z there's a r>0 such that there's an open ball contained with radius r in Z"

I read the proof about "An open ball is an open set" What I don't understand is: "what if we take a radius so big such that the open ball isn't completely contained in Z"

I just want to understand what I'm getting wrong.

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  • $\begingroup$ The definition you refer to seems to be for open sets in a metric topology. There are more general kinds of topologies. Any topology provides a family of open sets. Is that what you are interested in learning about? Real analysis starts with the "usual" metric topology on the real numbers. $\endgroup$ – hardmath Aug 21 '19 at 0:38
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    $\begingroup$ We don’t need all balls around $z$ to be in $Z,$ only some ball around $z.$ so the fact that the condition doesn’t work for large $r$ is irrelevant. You also don’t need one $r>0$ that works for all $z.$ $\endgroup$ – Thomas Andrews Aug 21 '19 at 0:41
  • $\begingroup$ That's what I thought, so if there is only one ball that is contained for that point then it satisfies the definition. $\endgroup$ – Max Villegas Aug 21 '19 at 0:44
  • $\begingroup$ Thank you Andrews $\endgroup$ – Max Villegas Aug 21 '19 at 0:47
  • $\begingroup$ "You also don’t need one r>0 that works for all z." @ThomasAndrews brings up a very good point. If you set is $(0,5)$ and the point is $3$ you need a ball of radius $2$ or smalller. If your point is $4.9$ you need a radius of $0.1$ or smaller. If your point $4.999999$ you nee a radius of $0.00000001$ or smaller. There is not radius that is small enough to work for all, but that doesn't matter. All that matters is that for every point there is a radius that works for that point. $\endgroup$ – fleablood Aug 21 '19 at 0:49
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... then you take a smaller radius.

A set $Z$ is open if for every point $x$ you can find an open ball (which might be very small) centered at $x$ that is contained in $Z$.

It doesn't mean every ball will work. It only means some balls will work. Actually it only needs to mean at least one ball will work. [That usually means the smaller balls will work too but sometimes there aren't any smaller balls (don't worry about that now.)]

If for example if you take the interior of a unit circle in a plane centered at $(0,0)$. That's open because "it has fuzzy edges"... Anyway if you take any point in it. Say $(0, 0.9999)$ then I can find an open ball centered at $(0,0.9999)$ entirely in the circle. To do that I have to take a radius $r \le 0.0001$; but I can do it if I take a radius small enough.

Now what if instead I had taken a radius of $27$ and that's way too big to fit in the circle? Well, I shrug my shoulders and give the person asking a confused look and say "Who cares about that radius? I can take a radius of $0.0001$ and that does work. I dont care about radii that don't work. I just care that there are radii the do work."

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  • $\begingroup$ why do you say that "sometimes there aren't any smaller balls..." ? I know this isn't the main point of your answer, but I'm curious as to what you meant $\endgroup$ – peek-a-boo Aug 21 '19 at 2:01
  • $\begingroup$ This is a nice explanation. But note that a ball centered at $(0,0.9999)$ with radius $r=0.0001$ is not in the interior of the unit circle. You need to choose $r{\color{red}{<}}0.0001$, so $r= 0.000099999999999$ will do the job. $\endgroup$ – Surb Aug 21 '19 at 9:06
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    $\begingroup$ @Surb an open ball contains the interior and not the edge. So, If I count the number of zeros correctly an open ball with radius $r=0.0001$ is in the interior of the unit circle as any point that is $0.0001$ away will not be a point of the open ball. $r = 0.0001$ is a correct radius (as will be all that are smaller). $\endgroup$ – fleablood Aug 21 '19 at 15:35
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By definition, a set $U$ in a metrix space $(X,d)$ is open if for every $x \in U$, there exists r>0 such that $B(x,r) \subseteq U$

In your case, in order to prove $B(x,r)$ is open, for every $y \in B(x,r)$, we choose a suitable $s>0$ for which $B(y,s) \subseteq B(x,r)$.

what if we take a radius $s$ so big such that the open ball isn't completely contained in the set ?

Why we need such a big $s$ ? By definition, we want there is such an s , so your case , $$s \leq r-d(x,y)$$ works!

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It seems that your specific question has been answered. It seems to me to be more important to point out that your version of the definition is wrong! If you want to understand a definition first you have to get the definition straight... (It may well be that what you meant by the definition is correct, but if so you didn't say what you meant.)

This is important - if you want to understand these things you need to be much more careful about the language. You say this:

Def 1. A set Z is open, if for every element z in Z there's a r>0 such that there's an open ball contained with radius r in Z.

Fixing up the English a little,

Def 1'. A set Z is open, if for every element z in Z there's a r>0 such that there's an open ball with radius r contained in Z.

This should look suspicious, because you say "for every element z in Z" but then you never mention z again in the rest of the definition.

The correct definition, probably (I hope) what you meant, is this:

Def 2. A set Z is open, if for every element z in Z there's a r>0 such that the open ball with radius r and center z is contained in Z.

Let's say the open ball with center $x$ and radius $r$ is $B(x,r)$, to make things easier to state. Then Def 2 is the same as

Def 2'. A set $Z$ is open if for every $z\in Z$ there exists $r>0$ such that $B(z,r)\subset Z$.

An example showing that the two definitions are not the same: Let's talk about suubsets of $\Bbb R$ with the standard metric $d(x,y)=|x-y|$. Define $$Z=(-1,1)\cup\{2\}=\{z\in\Bbb R:|z|<1\text{ or }z=2\}.$$Then $Z$ is not open, according to the correct definition, because if we say $z=2$ then $z\in Z$ but there does not exist $r>0$ with $B(z,r)\subset Z$.

But $Z$ does satisfy Def 1. First, $B(0,1)=(-1,1)\subset Z$. So the statement

(i) "there's a r>0 such that there's an open ball contained with radius r in Z"

is true (proof: let $r=1$).

And since (i) is true, it follows that

(ii) "for every element z in Z there's a r>0 such that there's an open ball contained with radius r in Z"

is true! (Because statement (i) simply doesn't mention $z$, the fact that (i) is true implies that (i) is true for every $z\in Z$.)

So Def 1 says that $Z$ is open, while Def 2 says it's not open.

If you followed that then at this point you're saying that's not what you meant. Fine, but that's exactly the important point: You need to be much more careful about the language or you have no chance with this advanced math stuff.

Note: It seems possible that English is not your native language. If so that's a valid excuse for giving Def 1 when it really makes no sense, should be Def 1'. But if you think that's an excuse for the whole thing you're fooling yourself! The difference between Def 1' and Def 2 or Def 2' is not just a matter of poor English expression, it's a matter of logic. Saying "for every z in Z" and then never mentioning z again makes no sense in any language.

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  • $\begingroup$ I didn't write it properly, I'm sorry. I'm gonna start writing in LaTeX. $\endgroup$ – Max Villegas Aug 21 '19 at 13:24
  • $\begingroup$ @MaxVillegas Using LaTeX would be better. But again, that's not the problem here - Def 1' is perfectly clear with no LaTeX, the problem is that it's just _wrong. Writing in LaTeX great, but much more important is to start being much more careful about definitions. (I decided not to reformat Def 1 in LaTeX lest we get the idea that was the problem. Look it up in the book: Def 1 is not what it says in the book. You thought it was the same - rephrasing thhings you don't quite understand is dangerous. ) $\endgroup$ – David C. Ullrich Aug 21 '19 at 13:41
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    $\begingroup$ Okay, I'll do better next time. $\endgroup$ – Max Villegas Aug 21 '19 at 13:44
  • $\begingroup$ @MaxVillegas Much better reply than the first one that you deleted! Why you got itt wrong is not the important issue. (If you say you have an audio recording or a photo of what was on the board fine - if you're just looking at your notes it seems more likely that what (s)he said was right and you just left out a few important words in your notes...) $\endgroup$ – David C. Ullrich Aug 21 '19 at 13:52
  • $\begingroup$ I wanted to write a reply but you're name wasn't on it when I submitted and I thought you wouldn't notice early. $\endgroup$ – Max Villegas Aug 21 '19 at 13:55

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