Question:
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?
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?
\[ E_1 + E_2 = \{x+y : x \in E_1,\ y \in E_2\} \] \[ E_1 \times E_2 = \{(x,y) : x \in E_1,\ y \in E_2\}. \] 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).
Updated On: Jul 21, 2026
- 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
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The Correct Option is C
Solution and Explanation
Step 1: Check option (A), sum with one set open.
Suppose \(E_1\) is open. Then \(E_1+E_2 = \bigcup_{y \in E_2} (E_1+y)\), a union over all \(y \in E_2\) of the translated sets \(E_1+y\). Translating an open set by a fixed vector gives another open set, since translation is a continuous, invertible map on a normed space. A union of open sets is open, so \(E_1+E_2\) is open. Option (A) is TRUE.
Step 2: Check option (B), sum of convex sets.
Let \(z_1 = x_1+y_1\) and \(z_2=x_2+y_2\) be in \(E_1+E_2\), with \(x_1,x_2 \in E_1\) and \(y_1,y_2 \in E_2\). For \(0 \leq \lambda \leq 1\),
\[ \lambda z_1+(1-\lambda)z_2 = \big[\lambda x_1+(1-\lambda)x_2\big] + \big[\lambda y_1+(1-\lambda)y_2\big]. \]
Since \(E_1\) is convex, the first bracket is in \(E_1\); since \(E_2\) is convex, the second bracket is in \(E_2\). So the whole point lies in \(E_1+E_2\), which shows \(E_1+E_2\) is convex. Option (B) is TRUE.
Step 3: Check option (D), product of connected sets.
It is a standard topology fact that the product of two connected spaces, with the product topology, is connected. Since \(E_1\) and \(E_2\) are connected, \(E_1 \times E_2\) is connected. Option (D) is TRUE.
Step 4: Check option (C), sum of closed sets, using a counterexample.
Work in \(X = \mathbb{R}^2\) with the usual norm. Let
\[ E_1 = \{(x,0) : x \in \mathbb{R}\}, \qquad E_2 = \left\{\left(x, \frac{1}{x}\right) : x > 0\right\}. \]
Both \(E_1\) (the \(x\)-axis) and \(E_2\) (one branch of a hyperbola) are closed subsets of \(\mathbb{R}^2\). A point of \(E_1+E_2\) is \(\left(x_1+x_2, \frac{1}{x_2}\right)\) with \(x_1 \in \mathbb{R}\), \(x_2>0\). Since \(x_1\) ranges over all reals, for any second coordinate \(v=\frac{1}{x_2}>0\) the first coordinate can take any real value, so \(E_1+E_2\) is exactly the open upper half plane \(\{(u,v): v>0\}\), which does not contain its boundary line \(v=0\) and so is not closed. Here two closed sets add up to a set that is not closed. Option (C) is FALSE, so this is the statement that is NOT TRUE.
Final Answer:
Options (A), (B), (D) are always true in a normed linear space, but (C) can fail, as the hyperbola-plus-line example shows. \[ \boxed{\text{Option (C)}} \]
Suppose \(E_1\) is open. Then \(E_1+E_2 = \bigcup_{y \in E_2} (E_1+y)\), a union over all \(y \in E_2\) of the translated sets \(E_1+y\). Translating an open set by a fixed vector gives another open set, since translation is a continuous, invertible map on a normed space. A union of open sets is open, so \(E_1+E_2\) is open. Option (A) is TRUE.
Step 2: Check option (B), sum of convex sets.
Let \(z_1 = x_1+y_1\) and \(z_2=x_2+y_2\) be in \(E_1+E_2\), with \(x_1,x_2 \in E_1\) and \(y_1,y_2 \in E_2\). For \(0 \leq \lambda \leq 1\),
\[ \lambda z_1+(1-\lambda)z_2 = \big[\lambda x_1+(1-\lambda)x_2\big] + \big[\lambda y_1+(1-\lambda)y_2\big]. \]
Since \(E_1\) is convex, the first bracket is in \(E_1\); since \(E_2\) is convex, the second bracket is in \(E_2\). So the whole point lies in \(E_1+E_2\), which shows \(E_1+E_2\) is convex. Option (B) is TRUE.
Step 3: Check option (D), product of connected sets.
It is a standard topology fact that the product of two connected spaces, with the product topology, is connected. Since \(E_1\) and \(E_2\) are connected, \(E_1 \times E_2\) is connected. Option (D) is TRUE.
Step 4: Check option (C), sum of closed sets, using a counterexample.
Work in \(X = \mathbb{R}^2\) with the usual norm. Let
\[ E_1 = \{(x,0) : x \in \mathbb{R}\}, \qquad E_2 = \left\{\left(x, \frac{1}{x}\right) : x > 0\right\}. \]
Both \(E_1\) (the \(x\)-axis) and \(E_2\) (one branch of a hyperbola) are closed subsets of \(\mathbb{R}^2\). A point of \(E_1+E_2\) is \(\left(x_1+x_2, \frac{1}{x_2}\right)\) with \(x_1 \in \mathbb{R}\), \(x_2>0\). Since \(x_1\) ranges over all reals, for any second coordinate \(v=\frac{1}{x_2}>0\) the first coordinate can take any real value, so \(E_1+E_2\) is exactly the open upper half plane \(\{(u,v): v>0\}\), which does not contain its boundary line \(v=0\) and so is not closed. Here two closed sets add up to a set that is not closed. Option (C) is FALSE, so this is the statement that is NOT TRUE.
Final Answer:
Options (A), (B), (D) are always true in a normed linear space, but (C) can fail, as the hyperbola-plus-line example shows. \[ \boxed{\text{Option (C)}} \]
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