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The first question is this example enter image description here

How can these intervals be open sets in $S$ when we can always take the points $0$ and $1$ and won't be able to find an open ball of radius $r>0$ whose points are all contained in these intervals?

The second question is in this partial proof of the theorem.

enter image description here

I don't see how if $A$ is open in $M$, then $S-A$ is closed in $S$. Here's how I tried to understand this.

Since $A$ is open in $M$, we can always choose a point in $ A\cap S$ and find a ball of radius $r>0$ completely contained in $A$. The intersection of this ball with $S$ is some region in $A\cap S$. If the original point we picked isn't on the boundaries of the region, we can always find a ball of radius $r_2$ small enough such that $r>r_2>0$ and the entire ball is contained in $A\cap S$ and so, $A\cap S$ is open in $S$. Now the set of elements in $S-A$ is the same as those in $S-(A\cap S)$ which shows that $S-A$ is closed in $S$. This argument only works if both $S$ and $A$ have no boundaries since we can't find a ball around a point on the boundary whose elements are all contained in the intersection, we know $A$ doesn't, but no such hypothesis was stated for $S$.

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  • $\begingroup$ a subset of $S$ is open in $S$ if and only if it is the intersection of $S$ with an open set in $\mathbb R$ $\endgroup$ – J. W. Tanner Aug 23 at 2:47
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    $\begingroup$ a better answer to the first question: an open ball of radius $r > 0$ is contained in the interval, since the open ball is with respect to the space $[0,1]$. For example, the ball of radius $\frac{1}{2}$ centered at $0$ is $[0,\frac{1}{2})$, since those are the points in $[0,1]$ that are within $\frac{1}{2}$ of $0$. $\endgroup$ – mathworker21 Aug 23 at 2:50
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    $\begingroup$ for your first question, the sets $[0, x) or (x, 1]$ are open in $S$ because of the definition. first, an open ball in a set $X$ $B(x; r) = \{z \in X: d(x, z)< r\}$, where $d$ is the metric function. Notice that $z \in X$, $z$ must be in X.Now. Now, a subset $G$ of $X$ is open if for all $t \in G$ there exists an $r>0$ such that $B(t, r) \subseteq G$, in other words, it's open if for any element of the subset, you can find a distance $r$ such that a ball with center on that element is totally contained in the set. Okay, so, for example the set $ A = [0, x) $, this set is open because $\endgroup$ – Donlans Donlans Aug 23 at 2:54
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    $\begingroup$ an open ball in that set of radius x is $B(x; r) = \{z \in A: d(x, z) < r \}$, in other words, it only takes elements that are in that set, therefore, for any element in that set, you can always find an $r$ greater than 0. if $t = 0$, $any r < x$ will do the job, because the only elements that its taking into account are those that are in $A$, if $t = 0$ and $r = x$, then the open ball would be $B(t;r) = \{z \in A: d(t,r ) < x\}$. even if,say for example, the distance between $0$ and $\frac{-1}{2}$ is less than $x$, it doesn't matter because $\frac{-1}{2}$ is not in the set $A$ $\endgroup$ – Donlans Donlans Aug 23 at 3:01
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    $\begingroup$ In the metric space $[0,1]$ then set $(1-r, 1]$ is an open ball with radius $r$. .... because $B(1,r) =\{y\in [0,1]: d(1,y) < r\} = \{y\in [0,1]: 1-r < y < 1+r\} = (1-r, 1+r)\cap [0,1] = (1-r,1]$. $\endgroup$ – fleablood Aug 23 at 3:17
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In a metric space $X$ with metric $d$, a subset $A$ of $X$ is open (relative to the metric) if and only if for every $a\in A$ there exists $\delta\gt 0$ such that $$B_X(a,\delta) = \{x\in X\mid d(x,a) \lt \delta\} \subseteq A.$$

Note that everything happens inside of $X$. You don’t get to “go outside” of $X$, even if such an outside naturally exists.

Now suppose you have a subset $S\subseteq X$. The metric of $X$ restricts to $S$, and so $(S,d|_S)$ is also a metric space, and hence we can ask about subsets of $S$ being open or not. When is $A\subseteq S$ open in this space? When for any $a\in A$, there exists $\delta\gt 0$ such that $$B_S(a,\delta) = \{x\in S\mid d(x,a)\lt \delta\}\subseteq A.$$ That is, the same condition, but now we only look at elements of $S$, not at elements of $X$. Why? because you aren’t looking at “open-in-$X$”, you are looking at open-in-$S$. You only look at elements of $S$, you ignore the elements that are not in $S$.

So when $X=[0,1]$ with the usual metric, the set $A=[0,x)$, $0\lt x\lt 1$, is open. Why? Because if we take $a\in A$, then $0\leq a\lt x$. Letting $\delta = \frac{1}{2}(x-a)$, we have that $$B_X(a,\delta) = \{ r\in [0,1)\mid d(a,r)\lt \delta\}$$ is completely contained in $A$.


Second question: $B$ is closed (in $M$), so its complement is open. $A=M-B$ is open, and then $B=M-A$ with $A$ open (in $M$).

Now, I claim that $S\cap A$ is open in $S$: if $a\in S\cap A$, then because $A$ is open in $M$, there exists $\delta\gt 0$ such that $B_M(x,\delta)\subseteq A$. Now consider $B_S(x,\delta)$. We have $$\begin{align*} B_S(x,\delta) &= \{s\in S\mid d(s,x)\lt \delta\}\\ &\subseteq \{s\in M\mid d(s,x)\lt \delta\} &\text{(since }S\subseteq M)\\ &= B_M(x,\delta)\\ &\subseteq A. \end{align*}$$ On the other hand, $B_S(x,\delta)\subseteq S$. Since $B_S(x,\delta)$ is contained in both $S$ and $A$, then $B_S(x,\delta)\subseteq S\cap A$.

Thus we have shown that for every $x\in S\cap A$, there exists $\delta$ such that $B_S(x,\delta)\subseteq S\cap A$. Hence $S\cap A$ is open in $S$.

Note that “boundary” doesn’t enter into it. This is all about balls in the appropriate ambient set.

Okay, now, since $S\cap A$ is open in $S$, then $S-(S\cap A)$ is closed in $S$. And now it is an easy check to verify that $$S-A = S-(S\cap A).$$ Thus, $S-A$ is closed in $S$, as claimed.

Again, note that boundary doesn’t enter into it.

Your error throughout is thinking that when you work in the relative metric of the subset, you should consider all points in the ambient space. You don’t; you just take points in the subset. Same way when we don’t take spheres in $\mathbb{R}^3$ when we look at open sets in $\mathbb{R}^2$.

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  • $\begingroup$ Thanks. Your answer is very clear. $\endgroup$ – Km356 Aug 23 at 3:32
  • $\begingroup$ @Km356: Now, here’s an interesting bit: what you are doing matters when going the other way. That is, if you start with $S\subseteq X$, and $A\subseteq X$, then “$A$ is open in $X$” implies “$A\cap S$ is open in $S$”. Now consider the other question: if $A\subseteq X$, does “$A$ is open in $S$” imply “$A$ is open in $X$”? Then the sort of thinking you were doing would come into play, and the issues of “boundaries” would actually matter. (The answer is that the implication is always valid if and only if $S$ is open-in-$X$) $\endgroup$ – Arturo Magidin Aug 23 at 3:38
  • $\begingroup$ I like how it turned out that my way of thinking isn't completely foolish :) $\endgroup$ – Km356 Aug 23 at 3:41
  • $\begingroup$ @Km356: It’s just not appropriate for the problem you were trying to apply it to. You needed to undo a screw, and you were trying to use ball hammer to do it. The ball hammer has its uses, just not that one... $\endgroup$ – Arturo Magidin Aug 23 at 3:43
  • $\begingroup$ @ArturoMagidin I don't understand these two sentences: "That’s because you aren’t looking at “open-in-𝑋”, you are looking at open-in-𝐴. You only look at elements of 𝐴, you ignore the elements that are not in 𝐴." It seems the sentence you said immediately before these two hit the right point, and then these two seem like gibberish to me, as of now. $\endgroup$ – mathworker21 Aug 24 at 0:32
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How can these intervals be open sets in S when we can always take the points 0 and 1 and won't be able to find an open ball of radius r>0 whose points are all contained in these intervals?

We will be able to.

What's the definition of "open ball"?

An open ball $B(x,r) = \{y \in X| d(y,x) < r\}$.

So if $X = [0,1]$ and then if $0 < r < 1$ then $B(1,r) = \{y\in [0,1]| d(y,0) <r\} = (1-r, 1]$.

That's an open ball. And $(x, 1]$ is an open set.

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I don't see how if A is open in M, then S−A is closed in S

If $A$ is open in $M$ then $A\cap S$ is open in $S$.

Pf: If $A$ is open in $M$ then every $x \in A\cap S \subset A$ will be such that there is an open ball with radius $r$ so that $B(x,r)=\{y\in M|d(y,x) < r\}$ so that $B(x,r) \subset A$.

But if $B(x,r) \subset A$ then $B_S(x,r) = \{y \in S|d(y,x) < r\}\subset A\cap S$. So $A\cap S$ is open in $S$.

ANd you had a theorem that if $O\subset X$ is open in metric space $X$ then $O^c = X\setminus O$ is closed in $X$.

So as $A\cap S$ is on in $S$, then the complement, $S \setminus A$, must be closed in $S$.

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