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Sorry if this is a stupid question, but this is confusing me somewhat. I have a homework question that requires me to find whether a subset (or subspace?) $A$ is $d$-open/$d$-closed and also whether it is $d_A$-open/$d_A$-closed. In my specific case I'm working with the Euclidean metric for $\mathbb{R}$, but I would like to understand generally as well as in this case.

I know when using an induced metric that it just inherits the properties of the metric of the whole space, and is only focused on $A$ and not the whole space. When I'm looking at it in terms of the metric of the whole space, I'm still looking at $A$ with a metric with the same properties, so I don't quite understand why I might come out with a different answer of whether it is open and closed compared to the induced metric.

I've been given the example:

Consider the open unit interval of real numbers $(0,1)$. This is not closed in $\mathbb{R}$ but it is closed in $(0,1)$.

Which makes sense, but I'm not sure how it applies to other examples such as my homework question in which the subset (or space?) is more complex than an interval.

I would greatly appreciate any explanation or other examples. Hopefully, I have communicated what I mean well enough.

Edit: The subset A in my case is $ A =\{\frac{1}{p} :p\in\mathbb{N} \text{ is prime}\}$ and the metric space is $(\mathbb{R}, d)$ where d is the Euclidean metric. I don't understand how it could be $d$-open and not $d_A$-open, aside from the interval example I already gave which was rather simple.

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    $\begingroup$ I suggest making your question more specific. This site works better with very specific questions. For instance, if you understand the open interval example, perhaps you can ask about the next most complicated example which you do not understand. $\endgroup$ – Lee Mosher Apr 24 '17 at 13:40
  • $\begingroup$ Ok, I will edit in my actual question. $\endgroup$ – Mike A Apr 24 '17 at 13:43
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    $\begingroup$ It applies just the same way: The distance between two points in $A$ is the same as their distance when viewed as pints in $\Bbb R$. However, whether a set is closed or open (or both or neither) does not depend on the distance between two points, but e.g. whether the limit of a sequenc of points is also in the set. $\endgroup$ – Hagen von Eitzen Apr 24 '17 at 13:43
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    $\begingroup$ You need to know what the open/closed sets are in the new (induced) topology. They are not the same as the original ones (check the definition). $\endgroup$ – AnyAD Apr 24 '17 at 13:44
  • $\begingroup$ @HagenvonEitzen Thanks for your response. So if I want to determine if $A$ is $d_A$-closed, I would see if there is a sequence with all its terms in $A$, but its limit point outside of $A$. If I wanted to determine if it is $d$-closed, what would I do differently in terms of looking for the sequence and its limit? Where would I look? $\endgroup$ – Mike A Apr 24 '17 at 13:54
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Let $A $ be a subset of $X $ where $(X,d) $ is a metric space.

Then a subset $S $ of $(A,d')$ ( where subspace matric $d'$ is restriction od $d $ to $A $) is open (in $(A,d')$) iff there is an open subset $V $ of $(X,d) $ such that $S=A\cap V$.

So if $A$ is open wrt ($\mathbb{R},d)$, then $A=A\cap A$ is open wrt $d_A$ by definition.

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    $\begingroup$ (For the OP): So for $B\subset A,$ the set $B$ may be open (or closed) in the space $A $ without being open (or closed) in the space $\mathbb R.$ For example if $A=\mathbb Q$ and $B=A\cap (-\sqrt 2\;,\sqrt 2)$ then $B$ is open and closed in $A$, $\endgroup$ – DanielWainfleet Apr 25 '17 at 4:55
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A subset $A$ of metric space $(X, d)$ is $d$-closed if for each $d$-convergent sequence with terms in $A$, the limit of the sequence is in $A$.

In contrast, a subset $A$ of metric space $(X , d)$ is $d_A$-closed only if it is an intersection with a $d$-closed set of $X$.

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