Prob. 25(a), Chap. 4 in Baby Rudin: In $\mathbb{R}^k$, the sum of a compact set and a closed set is closed

Here is Prob. 25 (a), Chap. 4 in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:

If $A \subset \mathbb{R}^k$ and $B \subset \mathbb{R}^k$, define $A + B$ to be the set of all sums $x+y$ with $x \in A$, $y \in B$.

(a) If $K$ is compact and $C$ is closed in $\mathbb{R}^k$, prove that $K+C$ is closed.

Hint: Take $z \not\in K+C$, put $F = z-C$, the set of all $z-y$ with $y \in C$. Then $K$ and $F$ are disjoint. Choose $\delta$ as in Exercise 21. Show that the open ball with center $z$ and radius $\delta$ does not intersect $K+C$.

Now here is Prob. 21, Chap. 4 in Baby Rudin, 3rd edition:

Suppose $K$ and $F$ are disjoint sets in a metric space $X$, $K$ is compact, $F$ is closed. Prove that there exists $\delta > 0$ such that $d(p, q) > \delta$ if $p \in K$, $q \in F$. Hint: $\rho_F$ is a continuous positive function on $F$.

Show that the conclusion may fail for two disjoint closed sets if neither is compact.

My effort:

We show that the set $\mathbb{R}^k \setminus (K+C)$ is open. Let $z$ be any point of $\mathbb{R}^k \setminus (K+C)$, and let the set $F$ be defined as $$F \colon= z-C = \left\{ \ z-y \ \colon \ y \in C \ \right\}.$$ Then the map $x \mapsto z-x$ of $\mathbb{R}^k$ into $\mathbb{R}^k$ is continuous, and under this map, the inverse image of the closed set $C$ equals $F$, which implies that $F$ is also closed in $\mathbb{R}^k$.

Now we show that the sets $K$ and $F$ are disjoint. Let $p$ be a point of $K \cap F$. Then as $p \in F$, so $p = z-y$ for some $y \in C$, and so $$z= p+y \in K+C, \ \mbox{ because } \ p \in K \ \mbox{ and } \ y \in C.$$ But $z \not\in K+C$. So $K$ and $F$ are disjoint.

Thus $K$ and $F$ are two disjoint sets in $\mathbb{R}^k$ such that $K$ is compact and $F$ is closed. So we can find a positive real number $\delta$ such that $$d(p, q) = \Vert p-q \Vert > \delta \ \mbox{ for every point p \in K and for every point q \in F}. \ \tag{1}$$

Now let $p \in \mathbb{R}^k$ such that $\Vert p-z \Vert < \delta$. If $p \in K+C$, then we note that $$p = x + y \ \mbox{ for some } \ x \in K \ \mbox{ and for some } \ y \in C.$$ Then $$p-z = (x+y) - z = x - (z-y).$$ Now as $x \in K$ and $z-y \in F$, so we must have $$\Vert x-(z-y) \Vert > \delta$$ by (1) above, which is the same as $$\Vert p-z \Vert > \delta,$$ and this contradicts the choice of $p$.

Thus we can conclude that if $p \in \mathbb{R}^k$ and $\Vert p-z\Vert < \delta$, then that $p$ cannot belong to $K+C$, which implies that $z$ is an interior point of $\mathbb{R}^k - (K+C)$.

But $z$ was an arbitrary point of $\mathbb{R}^k - (K+C)$. So $\mathbb{R}^k - (K+C)$ is open, and therefore $K+C$ is closed.

Is this proof correct? If so, then is my presentation clear enough too? If not, then where do problems lie?

• You are absolutely correct. – астон вілла олоф мэллбэрг Apr 7 '17 at 13:03
• @Masacroso thank you for your valuable comment, but can you please post a detailed enough answer as to how to use this approach? To be honest, I'm not sure how to show that a compact set in $\mathbb{R}^k$ is a Cartesian product of compact sets in $\mathbb{R}$. And, even if we do manage to show this, then exactly how to arrive at our desired conclusion. – Saaqib Mahmood Apr 7 '17 at 13:05
• @Masacroso: it is not true that an arbitrary compact set in $\mathbb R^2$ is a Cartesian product of compacts. And it is not true that the sum of closed sets is necessarily closed. – Martin Argerami Apr 7 '17 at 13:13
• @Martin oh, right. My bad. Not every subset of $\Bbb R^k$ can be defined by a cartesian product. – Masacroso Apr 7 '17 at 13:28