Question

I was wondering if you would be able to give me a solution or hint on the following problem:

Let D be a non-empty subset of the real numbers, and E={ax, x in D} where a>0. Prove that E is open if and only if D is open.

Any help would be greatly appreciated. Thank you

Answer 1

Question 1. Let $D$ be a non-empty subset of the real numbers, and $E = \{ax \mid x \in D\}$ where $a > 0$. Prove that $E$ is open if and only if $D$ is open.

Solution. It is sufficient to prove the implication: if $D$ is open, then $E$ is open (the converse implication is proved by applying the result for $\frac{1}{a} > 0$, because $D = \left\{\frac{x}{a} \mid x \in E\right\}$).

Let $x_0 \in E$, i.e. $x_0 = ay_0$ for some $y_0 \in D$. Since $D$ is open, there is $\varepsilon > 0$ such that all $y$ satisfying $y_0 - \varepsilon < y < y_0 + \varepsilon$ belong to $D$. Then $ay \in E$ for all such $y$. Note that $x_0 - a\varepsilon < ay < x_0 + a\varepsilon$ and, moreover, any $z$ from the interval $(x_0 - a\varepsilon, x_0 + a\varepsilon)$ can be expressed as $z = ay$, where $y \in (y_0 - \varepsilon, y_0 + \varepsilon) \subseteq D$. Thus, $(x_0 - a\varepsilon, x_0 + a\varepsilon) \subseteq E$, which proves that $E$ is open.

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