# Could someone explain the concept of a set being “open relative” to another set?

My professor sort of skimmed through this concept, giving only the definition and one example (i.e. $(0,1) \subset \mathbb{R}$ vs. $(0,1) \subset \mathbb{R}^2$, where $(0,1)$ is open relative to $\mathbb{R}$ but not $\mathbb{R}^2$), but no real explanation. I understand, more or less, this particular example, but I'm having trouble understanding this more generally.

Can somebody please intuitively explain this concept in the context of a metric space? (If it matters, we are using Rudin's Principles Of Mathematical Analysis).

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In order to know what "open relative to" means, you have to first know what "open" means:

A subset S of a metric space X is called open if, for every point p of S, there is a positive real number $\epsilon$ such that every point in X of distance less than $\epsilon$ of p lies in S.

Now, if we're working in some metric space and consider a subset A of that space, you can, if you wish, disregard the rest of the space and think of A as a metric space in itself. The notion of "open relative to A" is what you get when you realize, looking at the above definition, that some subsets of A that are not open subsets of the larger space may still be open subsets of A.

For example, suppose you look at the interval I=[0,1] as a metric space in itself, and disregard the rest of the real line. Then the subset S=(0.5,1] may seem at first not to be an open set, because you think, "Hey, there are points arbitrarily close to 1 that are not in S, so 1 is actually a boundary point!" Well, yes... except that none of those arbitrarily-close points are actually in I, so from the point of view of I as a space in itself, they don't exist. Every point in I sufficiently close to 1 is also in S, so if we want to work in I alone, we have to concede that S is open.

The moral of this story is that sets aren't inherently open: the definition depends on what metric space they are considered as subsets of. If you are only talking about one space it's harmless to just write "open," but otherwise you may have to be more specific. If you are considering both $\mathbb{R}$ and I, for example, writing "(0.5,1] is open" is ambiguous. So instead you write "(0.5,1]" is open relative to I."

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Thank you very much! –  Bachmaninoff Oct 12 '12 at 2:18
This is a great answer. –  Don Larynx Oct 30 '13 at 6:06
Two years later, still very useful in my understanding. Thank you. –  mathjacks Oct 30 '14 at 12:46

We say that a set $S$ is open relative to $A$ if for all $x\in S$, we can find an open ball (open with the topology/metric of $A$) centered at $x$ that is contained in $S$, i.e. there exists $\epsilon>0$ such that $B(x,\epsilon)\subset S$.

For your example, take $(0,1)$ in $\mathbb{R}$. The open balls in $\mathbb{R}$, with standard metric are the open intervals. Take any point in $(0,1)$, you can always find $\epsilon$ such that $0<|x-\epsilon|<1$, because of the density of $\mathbb{R}$.

Now take the same subset in $\mathbb{R}^2$. The open balls are now the open disks of the form $$(x-a)^2+(y-b)^2<r^2,$$ that is, an open disk of radius $r$ centered at $(a,b)$. Now for any point $(x,y)$ in $(0,1)$, a line segment, and for any $\epsilon$, the open ball $B(x,\epsilon)$ is going to leave the line segment. To see it, any point in $(0,1)$ will be of the form $(x,0)$ (or $(0,y)$), and so a disk centered at these kind of point is bound to have a $y$ coordinate (or a $x$) other than $0$, leaving it impossible for $B$ to be in $(0,1)$ entirely.

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Let $X$ denote a topological space. Let $A$ denote a subset of $X$. The set $B\subset A$ is open in $A$ if $B= A \cap U$ for some open set in $X$. When $X ={\mathbb R}$ then $B$ is a union of open intervals intersected with $A$. For example, suppose that $A$ denotes the rational numbers between $0$ and $1$. The an open interval in $A$ is an open interval of rationals. If $A$ is a closed interval, then an open interval is either open in the interior, or it is a half open interval in $A$. I specify intervals in these examples, because an open set of reals is a union of open intervals.

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If you have a topological vector space $V$ and sets $U \subset T \subset V$ you call $U$ open relative to $T$ iff $U$ is open in the subspace topology on $T$. Put differently, $U$ is open relative to $T$ if there is an open set $O \subset V$ with $U = O \cap T$.

The subspace topology is what you get if you restrict the topology of $V$ to $T$, i.e. the open sets of $T$ with the subspace topology are $\mathcal{O}_T = \{O \cap T : O \text{ open as subset of }$V$\}$. If you're dealing with a metric space, i.e. a topological vector space whose topology is induced by a metric, you get the same topology if you restrict the metric to $T$, and use that restricted metric to define a topology on $T$.

In your example, $U := (0,1)\times\{0\}$ is open relative to $T := \mathbb{R}\times\{0\}$, because $U = ((0,1)\times(0,1)) \cap T$ and $(0,1)\times(0,1)$ is open in $\mathbb{R}^2$. It isn't open as a subset of $\mathbb{R}^2$ because it doesn't contain an $\epsilon$-Ball around all of its members. The way you wrote that example (i.e, $(0,1)$ instead of $(0,1)\times\{0\}$), btw, doesn't make much sense because $(0,1)$ isn't even a subset of $\mathbb{R}^2$.

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Hey, thank you for the response. You're absolutely right, and I'm glad you knew what I meant! –  Bachmaninoff Oct 12 '12 at 2:15