Sum of two sets Let $P$ and $Q$ be two non-empty subsets of a normed linear space $X$ and let $$P+Q=\{x+y\in X: x\in P\mbox{ and }y\in Q\}.$$ 
Then which of the following is FALSE?


*

*If $P$ or $Q$ are convex, then $P+Q$ is convex.

*If $P$ or $Q$ is open, then $P+Q$ is open.

*P+Q must be closed if P and Q are closed.

*If $P$ is closed and $Q$ is compact, then $P+Q$ is closed.


I think 1st is false, because if one set is not convex, then on addiction with a convex set, the resultant might not be convex.
But in one answer key, the answer is given as 3rd. In another, the answer is given as 4th. But I am not able to think of counter examples for either 3rd or 4th.

1 and 3 are false and the other two are true.
Counterexample for 1 is $P=[0,0.5]$ and $Q=[1,2]\cup[3,4]$, then $P+Q=[1,2.5]\cup[3,4.5]$, which is non-convex.
Counterexample for 3, as pointed out, is $P=\{-n+\frac{1}{n}:n\in N \}$
and $Q=N$, then $P+Q$ has subset $X=\{\frac{1}{n}:n\in N \}$ which has limit point at $0$ but $0 \notin P+Q$. 
Proof of 2 is as follows:
$$P+Q=\cup_{q\in Q} (P+q)$$
Now $P+q$ is open set is easy to see and the fact that union of open set is open gives us our result.
For 4 refer this Proof check of sum of a compact and closed set of real numbers is closed 
