# Is the sum of two compact subsets of Rn compact + another problem

• Sep 20th 2010, 06:02 PM
BBAmp
Is the sum of two compact subsets of Rn compact + another problem
Hello everyone,

1) I have been banging my head against the wall trying to figure out how to prove that two subsets $A, B$ which are compact subsets of $R^n$ produce another compact subset when added together.

The problem is.. I know the definition of a compact set, but is that the right place to start? Can anyone please gently guide me in the right direction?

2) if $U$ is an open subset in Rn and V is an arbitrary subset, then U+V is open. Here I am having issues because again I know the definition of an open subset but I become paralyzed when I try to do anything with it. Frankly I'm not sure if V is an arbitrary subset of Rn but even if I assume it is, how on earth do I start?

I am still very new to proofs and I would like to learn on my own so please post only ideas of where I can or should start.

Thank you.
• Sep 20th 2010, 07:29 PM
BBAmp
okay this is how i went about solving the first one.. I would really appreciate input on if my thought process is correct since I am learning how to write proofs.

since A and B are compact subsets of Rn, A and B each have open covers where $(O_{\alpha})_{\alpha\epsilon scriptA}$ is an open cover of A (where scriptA is some index set), and $(P_{\beta})_{\beta\epsilon scriptB}$ is an open cover of B (where scriptB is another index set). Then there are $\alpha_{1},...,\alpha_{m} \epsilon scriptA$ and $\beta_{1},...,\beta_{m} \epsilon scriptB$ such that $A \subseteq \cup O_{\alpha}$ and $B \subseteq \cup P_{\beta}$ which comes from the definition of compactness.

Adding A and B creates a new set C. In order for C to be compact, it must have its own set of open covers where the union of a certain collection of them (much like the union of open covers that contain A and B above) where $C \subseteq \cup Q_{\gamma}$. That means:

$C = (A+B) \subseteq ( \cup O_{\alpha} + \cup P_{\beta})$

then since $C = A+B \Rightarrow C \subseteq (\cup O_{\alpha} + \cup P_{\beta})$
also since $C = A+B \Rightarrow A+B \subseteq (\cup Q_{\gamma})$

which implies $\cup O_{\alpha} + \cup P_{\beta} = \cup Q_{\gamma}$ so c is compact.

• Sep 26th 2010, 04:27 AM
HallsofIvy
Quote:

Originally Posted by BBAmp
okay this is how i went about solving the first one.. I would really appreciate input on if my thought process is correct since I am learning how to write proofs.

since A and B are compact subsets of Rn, A and B each have open covers where $(O_{\alpha})_{\alpha\epsilon scriptA}$ is an open cover of A (where scriptA is some index set), and $(P_{\beta})_{\beta\epsilon scriptB}$ is an open cover of B (where scriptB is another index set). Then there are $\alpha_{1},...,\alpha_{m} \epsilon scriptA$ and $\beta_{1},...,\beta_{m} \epsilon scriptB$ such that $A \subseteq \cup O_{\alpha}$ and $B \subseteq \cup P_{\beta}$ which comes from the definition of compactness.

This is an incorrect definition of "compactness". Saying that a set is compact does NOT mean that there exist open covers that have finite subcovers- it means that every open cover has a finite subcover. In particular, that means that you cannot assume that an open cover of A+ B is constructed from given open covers of A and B.

By the way, how are you defining A+ B for A and B in $\mathbb{R}^n$? $\{u+ v| u\in A, v\in B\}$ and "a+ b" is coordinatewise addition?

Quote:

Adding A and B creates a new set C. In order for C to be compact, it must have its own set of open covers where the union of a certain collection of them (much like the union of open covers that contain A and B above) where $C \subseteq \cup Q_{\gamma}$. That means:

$C = (A+B) \subseteq ( \cup O_{\alpha} + \cup P_{\beta})$

then since $C = A+B \Rightarrow C \subseteq (\cup O_{\alpha} + \cup P_{\beta})$
also since $C = A+B \Rightarrow A+B \subseteq (\cup Q_{\gamma})$

which implies $\cup O_{\alpha} + \cup P_{\beta} = \cup Q_{\gamma}$ so c is compact.