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Given $x_1,x_2 \in \{y \in \mathbb{R}^n: y\ge0, 1^Tx=1\}$.

I have the set $S=\{0 \le q\le 1, x_1^Tq \le c\}$ where $c \in [0,1]$ and the inequalities are to be understood component-wise.

The set $S_1=\{q \in \mathbb{R}^n: 0\le q \le 1\}$ is compact, and I think S2 = $\{q \in \mathbb{R}^n: x_1^Tq \le c \}$ is also a compact set because $x_1 \ge 0$, and $0\le x_1^Tq \le c$, i.e. $x_1$ is in the non-negative hyper-octant, and $S$ is sandwiched between two hyperplanes.

I want to know whether I have a compact set.

My understanding is that I have two compact sets. Given the intersection of two compact sets, can I say that I have a compact set?

pippp
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1 Answers1

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The terminology "compact set over a variable" does not make any sense.

Anyway, if I am interpreting your question correctly (i.e. $q$ is an identifier for an arbitrary point in $\mathbb{R}^n$), then yes, the finite intersection of compact sets is again compact, so in particular the intersection of two compact sets is a compact set.

As for the specific sets in question, $S_1$ looks like the unit cube (assuming you are using inequalities here in the component-wise sense), but I don't see why $S_2=\{q \in R^n: x_1^T q \le c \}$ would be compact; this looks like a half-space to me, which would be unbounded and hence not compact.

Although ultimately that doesn't matter, since $S_2$ is closed, and the intersection of a closed set and a compact set is again compact (at least in a Hausdorff space like $\mathbb{R}^n$).

Chill2Macht
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  • Thanks for comments. Here, x1 is defined as simplex. I was thinking about how I can use this fact about simplex. I guess it is not necessary to consider the fact that x1 is simplex? I understood S2 is closed half space (= closed set). However, I did not follow why the intersection of a closed set and compact set is compact. Can you provide any further comments or reference on this? – pippp Apr 30 '16 at 18:01
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    Well specifically, and closed subset of a compact space is again compact. Since the intersection of a closed space and a compact space is closed (since the compact space is closed), it is a closed subspace of a compact space, and hence compact. See for example: https://proofwiki.org/wiki/Closed_Subspace_of_Compact_Space_is_Compact – Chill2Macht Apr 30 '16 at 18:10
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    http://planetmath.org/closedsubsetsofacompactsetarecompact, https://math.dartmouth.edu/archive/m54x12/public_html/m54lecture15.pdf, http://math.stackexchange.com/questions/212181/a-subset-of-a-compact-set-is-compact, https://en.wikibooks.org/wiki/Topology/Compactness, etc. – Chill2Macht Apr 30 '16 at 18:11
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    When you say $x_1$ is defined as a simplex, you mean the set $S_1$ is supposed to be a simplex? Luckily the standard (n-1)-simplex is also compact, not just the unit cube, i.e. the argument I gave above still holds. The notation for the standard (n-1)-simplex would be ${q \in \mathbb{R}^n: \sum\limits_{i=0}^n t_i =1, t \ge 0 }$. I thought this looked like a linear programming problem, hence why I thought to interpret the inequalities component-wise. – Chill2Macht Apr 30 '16 at 18:15
  • Thank you for further detail comments. I meant that x1 from S2 is defined as a simplex. Given x1, simplex, do you think that I can just ignore the fact that x1 is simplex and consider S2 is closed half space. Therefore the entire set become compact set? – pippp Apr 30 '16 at 20:09
  • how can a point $x_1$ be a simplex? A simplex is a set, not a point? – Chill2Macht Apr 30 '16 at 20:21
  • Please check my modified questions again (updated the set S). I meant that x1 and x2 are n dimensional vectors that are elements of simplex. – pippp Apr 30 '16 at 21:19
  • Ok so $x_1$ and $x_2$ are elements of the (n-1)-dimensional simplex, i.e. probability distributions on {1,...,n}. I don't understand how you have re-written the definition of $S$ whatever, or what exactly the question is now? The notation ${x \in A: ... }$ means "the set of all elements $x$ in the set $A$ such that the property "..." holds. The way $S$ has been re-written, it is no longer clear which expression is supposed to be the elements of the set, and which expressions are supposed to be the defining conditions that a point needs to satisfy in order to be a member of the set. – Chill2Macht Apr 30 '16 at 22:07
  • If you are asking whether or not the closed half-space $S_2$ is compact, then no, $S_2$ is not compact, it is only closed. The Heine-Borel Theorem tells us that a set in $\mathbb{R}^n$ is compact if and only if it is closed and bounded -- the closed half-space $S_2$ is closed, but not bounded (it is by definition unbounded), hence not compact. – Chill2Macht Apr 30 '16 at 22:15
  • See: https://en.wikipedia.org/wiki/Heine%E2%80%93Borel_theorem – Chill2Macht Apr 30 '16 at 22:22
  • I'm sorry for confusion. I included objective function in the set, so I made you confused. I guess when I want to check whether the set S is compact, I do not have to consider the property of x1. Closed halfspace with compact set will be compact set based on the Heine Borel Theorem because compact set gives bounds for unbounded closed halfspace? – pippp Apr 30 '16 at 22:53
  • Yes exactly. Their intersection will be closed and bounded, hence compact. – Chill2Macht May 01 '16 at 00:32