Let \(E_1\) and \(E_2\) be subsets of a normed linear space \(X\), and \[ E_1 + E_2 = \{x+y : x \in E_1,\ y \in E_2\} \] \[ E_1 \times E_2 = \{(x,y) : x \in E_1,\ y \in E_2\}. \] Which one of the following is NOT TRUE?
Show Hint
Check each claim directly: open-plus-anything and convex-plus-convex are provable from definitions, and connected times connected is a standard fact, but closed-plus-closed has a classic counterexample (a line and a hyperbola branch in the plane).
If either of \(E_1\) or \(E_2\) is open, then \(E_1+E_2\) is open.
If \(E_1\) and \(E_2\) are convex, then \(E_1+E_2\) is convex.
If \(E_1\) and \(E_2\) are closed, then \(E_1+E_2\) is closed.
If \(E_1\) and \(E_2\) are connected, then \(E_1 \times E_2\) is connected.
Show Solution
The Correct Option isC
Solution and Explanation
To find which statement is NOT true, it helps to know that three of these four properties are standard preserved-property results for normed linear spaces, and one is a classic trap. Let's go through each option.
(A) open plus anything gives an open sum: if $E_1$ is open, then for each fixed $y \in E_2$ the shifted set $E_1+y$ is open, because adding a constant vector is a continuous, invertible map on a normed space. Since $E_1+E_2 = \bigcup_{y \in E_2}(E_1+y)$ is a union of open sets, it is open. This holds true.
(B) convex plus convex is convex: for $z_1=x_1+y_1$, $z_2=x_2+y_2$ in $E_1+E_2$ and $\lambda \in [0,1]$, the point $\lambda z_1+(1-\lambda)z_2$ splits as $[\lambda x_1+(1-\lambda)x_2] + [\lambda y_1+(1-\lambda)y_2]$, a sum of a convex combination inside $E_1$ and one inside $E_2$, so it lands back in $E_1+E_2$. This holds true.
(D) connected times connected is connected: this is a basic topology fact, the product of connected spaces with the product topology is always connected. This holds true.
(C) closed plus closed is closed: this is the one that can fail. Take $X=\mathbb{R}$, $A=\{0,1,2,3,\ldots\}$ (the non negative integers, a closed discrete set) and $B=\{-n+\frac{1}{n} : n=1,2,3,\ldots\}$ (also closed, since its points march off to $-\infty$ with no finite accumulation point). Taking the element $n$ from $A$ and $-n+\frac{1}{n}$ from $B$ with matching index gives the sum $\frac{1}{n}$, which tends to $0$ as $n \to \infty$ but is never equal to $0$ for any finite $n$. So $0$ is a limit point of $A+B$ that does not belong to $A+B$, meaning $A+B$ is not closed even though both $A$ and $B$ are closed.
So three of the four statements are guaranteed by general theorems, and only the closed-plus-closed claim can break down, as the counterexample above shows.
Let's summarize:
Open-plus-anything, convex-plus-convex, and connected-times-connected are always safe, provable directly from definitions.
Closed-plus-closed is the classic exception: two closed sets can add up to a non closed set, as the integers and the shifted-reciprocal sequence show.
