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If $A\subset B$ and $B$ is a metric space. What is the difference between "$A$ open subset of $B$" and "$A$ open relative to $B$"?

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A open relative to B is often used to express the fact that A is open in the subspace topology This would howerver require a slightly different setting than you described - you would need a larger space which contains B. – Martin Sleziak Jul 17 '11 at 12:02

There is only a difference if $B$ itself is a subspace of some larger (say metric) space $X$: i.e., $A \subset B \subset X$.

Then, "$A$ is an open subset of $B$" means that (i) $A \subset B$ and (ii) $A$ is open in $X$. However "$A$ is open relative to $B$" means that (i) $A \subset B$ and (ii) $A$ is open when viewed as a subspace of $B$, i.e., given any $x \in A$, there exists an $\epsilon > 0$ such that every element of $B$ of distance less than $\epsilon$ from $x$ lies in $A$. (However it need not be the case that every element of X with distance at most $\epsilon$ from $x$ lies in $A$.)

Here is a simple example of this: suppose that $X = \mathbb{R}^2$, $A = (0,1) \times \{0\}$ and $B = [0,1] \times \{0\}$ -- that is, $A$ is an open interval on the horizontal line $y = 0$ and $B$ is the corresponding closed interval. Then $A$ is open relative to $B$ but is not an open subset of $B$.

(Note though that $B$ is not just closed in itself -- which is trivial -- or closed in $\mathbb{R} \times \{0\}$ but is actually closed in $X = \mathbb{R}^2$. Not coincidentally, $B$ is compact. Compactness can be viewed -- among other ways -- as a sort of "absolute closedness": if a subset $X$ of a metric space $Y$ is compact, then it is not only closed in $Y$ but in fact in every metric space $Z$ containing $Y$ as a subspace.)

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+1 A nice answer (as usual). However, there is a "one-dimensional example": $[0,\frac{1}{2})$ is open relative to $[0,1]$ but is not an open subset of $[0,1]$. – Amitesh Datta Jul 17 '11 at 12:39
@Amitesh: sure. I chose the example I did because $A$ "looks open" but wanted to emphasize that this depends on where it is viewed. If memory serves, Rudin gives a similar example early on in his Principles of Mathematical Analysis. – Pete L. Clark Jul 17 '11 at 13:06

If $(X,\rho)$ is a metric space, then we say that $A\subseteq X$ is an open subset of $X$ if to each $a\in A$, there exists $\delta>0$ such that $b\in A$ whenever $\rho(a,b)<\delta$.

If $(X,\rho)$ is a metric space and if $Y\subseteq X$ is a subspace of $X$, then we say that $A\subseteq Y$ is open relative to $Y$ if to each $a\in A$, there exists $\delta>0$ such that $b\in A\cap Y$ whenever $\rho(a,b)<\delta$ and $b\in Y$.

Exercise 1: If $X$ is a metric space and if $A\subseteq X$, prove that $A$ is an open subset of $X$ if and only if $A$ is open relative to $X$.

Exercise 2: Let $X$ be a metric space and let $Y\subseteq X$. Prove that if $A$ is an open subset of $X$, then $A\cap Y$ is open relative to $Y$.

Exercise 3: Let $X$ be a metric space and let $Y\subseteq X$. If $A\subseteq Y$, then is it true that $A$ is an open subset of $X$ if $A$ is open relative to $Y$? Prove or give a counterexample.

Exercise 4: Let $X$ be a metric space and let $Y$ be an open subset of $X$. Prove that if $A\subseteq Y$ is open relative to $Y$, then $A$ is an open subset of $X$.

Exercise 5: Let $X$ be a metric space and suppose $X=\bigcup_{n=1}^{\infty} X_n$. If $A\subseteq X$ is such that $A\cap X_n$ is open relative to $X_n$ for all positive integers $n$, then is it true that $A$ is an open subset of $X$? Prove or give a counterexample.

Research Project: Let $\{X_{\alpha}\}_{\alpha\in A}$ be a collection of subspaces of the metric space $X$. We say that the topology of $X$ is coherent with the subspaces $\{X_{\alpha}\}_{\alpha\in A}$ if the following property is satisfied:

A subset $A$ of $X$ is open in $X$ if and only if $A\cap X_{\alpha}$ is open relative to $X_{\alpha}$ for all $\alpha\in A$.

Investigate thoroughly necessary and sufficient conditions for the topology of $X$ to be coherent with a collection of subspaces. More specifically, state as many results and examples as you can in this connection. After you have thought about this problem sufficiently deeply, you may also refer to the internet and textbooks for further discussion.

Warning: Solutions to Exercises

Solution to Exercise 1: Since $A\cap X=A$, a careful examination of the definitions above shows that $A$ is an open subset of $X$ if and only if $A$ is open relative to $X$.

Solution to Exercise 2: Let $a\in A\cap Y$. Since $A$ is an open subset of $X$, there exists $\delta>0$ such that $b\in A$ whenever $\rho(a,b)<\delta$. In particular, we have $b\in A\cap Y$ whenever $\rho(a,b)<\delta$ and $b\in Y$. Therefore, $A\cap Y$ is open relative to $Y$.

I hope this helps!

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Could anyone please elaborate on how I can make this answer more useful? I am just wondering since my answer has not (at this point in time) received any upvotes. (I know Pete's answer has and it is an excellent answer but I am wondering what I need to do to make my answer good as well.) – Amitesh Datta Jul 17 '11 at 12:42
There is a big room for personal philosophy and perhaps a discussion in the meta. In my (oh so humble) opinion you phrase your answer excessively as exercises to the reader. While it is a good practice to prove things on your own, nobody likes books that provide you with no actual proof but rather a series of hints and directions. You should perhaps give some partial answers to some of the exercises (I have noticed that you do this quite often, and I think that this is a matter of taste, but you should perhaps start a meta.MSE discussion on the topic. Also, I upvoted your answer.) – Asaf Karagila Jul 17 '11 at 16:16
@Amitesh: I think it may be good to point out that how the exercises are related to the OP. – Jack Jul 17 '11 at 16:35
@Amitesh I hope that most visitors appreciate the work you put into constructing these small assignments --- in general I think such posts are more helpful than ones which give away the entire game immediately. I didn't upvote this one when I saw it simply because it's about six times more involved than the original question, and I didn't have time to look the whole thing over. – Dylan Moreland Jul 17 '11 at 23:38
Dear Amitesh, you showed no mercy... I fear that this was time badly invested. You were not rude at all and you spelled out exactly what I thought, but I decided to leave it at a remark on the rather inappropriately condescending way this "answer" was phrased, and dismiss the matter with a "I don't understand the rest of your points". I hope Mathemagician1234 will look closely at your comments and learn something from your dedicated dissection of his "points", but I'm afraid that it won't happen. – t.b. Jul 18 '11 at 8:47

I just came across this definition in Principles of Mathematical Analysis and perhaps this might be helpful to someone as I had to read it over a couple of times and even though this is minor, it helped me.

A small notational adjustment to the definition of the open ball:

Definition: Let $X$ be a metric space and $p \in X$ then let $N_r^X(p) := \{q \in X \mid d(p,q) < r\}$.

Now, using that $E$ is open if every point of $E$ is an interior point of $E$ which is Definition 2.18, I prefer this way of defining openness:

Definition: A set $E \subset X$ is open in $X$ if for each $p \in E$ there is an associated $r > 0$ such that $N_r^X(p) \subset E.$

So using the above we now get:

Definition: A set $E \subset Y$ is open relative to Y if for every $p \in E$ there is an associated $r >0$ such that $N_r^Y(p) \subset E.$

Why do I find this helpful? Well, using the above and working an example you can show that an open ball isn't necessarily a "ball" anymore and I think it's very helpful to realize that quickly as it tends to highlight how our usual notation of open/closed sets are relative.

For example, take $X = \mathbb{R}, Y = [0,1] \text{ and } E = [0,\frac{1}{2})$. The open ball $N_r^Y(0)$ turns out to be $[0,r)$ and is no longer $(-r,r)$ as we tend to expect it to be. So in this specific example we can readily see that by setting $r =\frac{1}{4}$ that $N_{1/4}^Y(0) = [0,\frac{1}{4}) \subset [0,\frac{1}{2}) = E $. So the "problematic point" turns out not to be a problem at all.

Hope this may help someone!

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+1 and thanks for the time in compiling this informative answer! – Amitesh Datta Aug 22 '13 at 6:28
@AmiteshDatta When I posted this answer I didn't know if anyone was ever going to read. Glad to see someone has. Thanks. – BU982T Aug 22 '13 at 10:41
Hi @Bryan, certainly, no problem, I'm glad I had the pleasure to read your answer! Unfortunately, a lot of excellent answers go unnoticed on this website and I hope I can do a small part to rectify this trend. I sometimes like to go through my past activity on this website (nostalgia?) and I stumbled on your answer by doing so. Ironically, it looks like I was complaining about my answer above (written two years ago! amazing how fast time goes by) not receiving upvotes when it received 7! Anyway, thanks again for your answer; we've got an excellent set of answers here! – Amitesh Datta Aug 22 '13 at 11:51

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