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I have just started with a course on convex optimization and have been introduced to the concept of the interior of a set. I have a fairly basic question. I am still trying to understand this topic, so please forgive me if this is a stupid or incorrect question.

Why is not the definition of the interior of a set not tied to the dimension of the space it is lying in?

For e.g. Consider a square in $\mathbb{R}^2$ i.e. $\mathcal{S} = \{(x,y)|x\in[-1,1],y\in[-1,1]\}$. By definition, points inside the square are interior points, because all points within an open '${disc}$' of radius $\epsilon>0$, around the point under consideration, lie inside the set.

But in 3 dimensions, this 2D figure has $\varnothing$ interior, since all points inside an open '$ball$' around any of the points inside $\mathcal{S}$, do not belong to $\mathcal{S}$

So, shouldn't the definition of the interior of a set be tied to the dimension of the space from which we are looking at the set? Else, how does one know whether to use a disc or a ball to find the interior of a set.

Thanks in advance!

Karthik Upadhya
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  • That would be the concept of relative interior ($\operatorname{ri}$), which is the interior relative to the affine hull of the set in question. Unfortunately $\operatorname{ri}$ has a few quirks that make it easy to trip on. For example, if $A \subset B$ it follows that $A^\circ \subset B^\circ$, but the same does not necessarily apply to $\operatorname{ri}$. – copper.hat Sep 22 '14 at 06:04
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    The definition of interior does depend on the space you're working in (in exactly the ways you mentioned). What definition have you seen? – ShreevatsaR Sep 22 '14 at 06:19
  • Better to speak not of spheres, but of open balls (sometimes called open discs). The sphere is just the surface of the ball; notice, for example, JackM's answer does not mention spheres. – goblin GONE Sep 22 '14 at 07:08
  • @ShreevatsaR Thank you for your answer. The definition was from the book Convex optimization, Boyd and from Wikipedia. Both of them mentioned an open sphere without mentioning the dimension. – Karthik Upadhya Sep 22 '14 at 07:48
  • @goblin Have made the edit. Thank you. – Karthik Upadhya Sep 22 '14 at 07:49

3 Answers3

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The definition of the interior is:

A point $x$ is in the interior of $S$ iff it can be surrounded by an open ball lying entirely inside $S$.

The reference to dimension is hiding inside the term open ball. In two dimensions, this is a disk, in three dimensions a literal ball, and so on.

Jack M
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  • Thank you for your answer. But my question is precisely this. If one is asked to find the interior of a set, what dimension should he assume? – Karthik Upadhya Sep 22 '14 at 07:51
  • @KarthikUpadhya, this is very hard to explain in a Q&A format, because its a subtle point, and mathematics is full of abuses of notation that make the point even more subtle. I really need to be there in person so we can have a conversation. This is the only way to explain it properly. But let me try anyway. You don't need to know the dimension.What you need is a change of perspective. The relevant perspective is this. – goblin GONE Sep 22 '14 at 08:14
  • Firstly, $\mathbb{R}^2$ is not a subset of $\mathbb{R}^3$; they should be imagined as disjoint. Secondly, $\mathbb{R}^2$ is not on equal footing with its subsets. A subset $S$ of $\mathbb{R}^2$ "remembers" that it is a subset of $\mathbb{R}^2$. On the other hand, $\mathbb{R}^2$ "floats free" in the universe. So if $S$ is a subset of $\mathbb{R}^2$, then $\mathrm{int}(S)$ is unambiguous; we're taking the interior in $\mathbb{R}^2,$ because $S$ is a subset of $\mathbb{R}^2$, because it remembers that it is a subset. Have a look at my answer again, especially the last paragraph. – goblin GONE Sep 22 '14 at 08:15
  • Also, @KarthikUpadhya, make sure you're familiar with the definition of a metric space, since this is the basic concept necessary to understand all the rest. – goblin GONE Sep 22 '14 at 08:19
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    @KarthikUpadhya If you ask me to find the interior of a set, I need to know what topology you're using. In other words, you need to tell me explicitly what dimension to use. – Jack M Sep 22 '14 at 10:53
  • @JackM Got it. Thank you! – Karthik Upadhya Sep 22 '14 at 11:02
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The idea of the interior is a concept from the world of topology. It is designed to work with things where you don't (yet) even have an idea of what dimension means. On the other hand, as mentioned in the comments, the definition is (intentionally) depended on what topological space you are looking at -- a square in $\mathbb{R}^2$ is distinct from a square in $\mathbb{R}^3$ because $\mathbb{R}^2$ is different from $\mathbb{R}^3$. To make sense of interior, you must consider the set of interest together with the whole space it is lying in.

Jessica B
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You're correct; the context in which interiors are taken matters.

It is a mathematical meteprinciple that the way to formalize "widening of context" is by leveraging the concept of an embedding. Nobody has (so far) proposed a definition of embedding that works in all cases, but the correct definition for metric spaces is surely that of an isometry, sometimes called an isometric embedding. (Note: isometries are always injective; however, contrary to what you might expect from the prefix iso-, they needn't be invertible, nor surjective.)

The importance of context for the taking of interiors can therefore be phrased as follows. Let $X$ and $Y$ denote metric spaces, and suppose $f : X \rightarrow Y$ is an isometry (i.e. an isometric embedding, in other words a widening of context). Then given a subset $S \subseteq X$, all we can say is that $f(\mathrm{int}(S)) \supseteq \mathrm{int}(f(S)).$ In particular, the reverse inclusion needn't hold.

Here's a simple example, a slight variant on the example you give. Bind the variables $u$ and $v$ to the real number line. Take $X = \mathbb{R}^2, \,Y = \mathbb{R}^3$, and let $f : X \rightarrow Y$ denote the function specified as follows. $$f(u,v) = (u,v,0)$$

(Check that this is an isometry; i.e. a widening of context.)

Now let $S \subseteq X$ be defined as follows. $$S=\{(u,v) \in X\,|\,u∈(−1,1),v∈(−1,1)\}$$

Then:

  1. $\mathrm{int}(S) = S,$ so $f(\mathrm{int}(S)) = \{(u,v,0) \in Y\,|\,u∈(−1,1),v∈(−1,1)\},$ which is not empty.

  2. On the other hand, we have $f(S) = \{(u,v,0) \in Y\,|\,u∈(−1,1),v∈(−1,1)\},$ so $\mathrm{int}(f(S)) = \emptyset_Y$.

So yes, context matters. It is considerations such as these that led to the by-now dominant way of looking at mathematical structures. If we're thinking of $\mathbb{R}^2$ and $\mathbb{R}^3$ as metric spaces, then we imagine them "floating free" in the universe. We do not imagine that $\mathbb{R}^2 \subseteq \mathbb{R}^3$. Rather, there are many embeddings $f : \mathbb{R}^2 \rightarrow \mathbb{R}^3$, and each of these gives us a way of viewing $\mathbb{R}^2$ as a subspace of $\mathbb{R}^3$. But these $f$'s are not harmless; as we've just seen, widening the context actually changes things; some constructions are preserved under embeddings, but many are not.

goblin GONE
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  • I loved your answer. Especially the last paragraph. I quite did not understand how you arrived at $f(\mathrm{int}(S)) \supseteq \mathrm{int}(f(S))$. – Karthik Upadhya Sep 22 '14 at 08:14
  • @KarthikUpadhya, to be honest, I can't quite see how to prove this. I'll edit when I think of the proof, if I do not forget. And thank you for the kind words. You're awesome. – goblin GONE Sep 22 '14 at 08:17