Let $\emptyset \ne A \subset \mathbb{R}$ and $-2A = \{-2x \mid x \in A \}.$ Show that $\inf(-2A) = -2\sup A$ 
Let $\emptyset \ne A \subset \mathbb{R}$ and $-2A = \{-2x \mid x \in A \}.$ Show that $$\inf(-2A) = -2\sup A$$

Is there a way to prove this using the $\epsilon$ definition for suprema and infima? If $\inf(-2A)= \beta$, then $\forall \varepsilon >0$ there exists $x_\varepsilon$ such that $x_\varepsilon < \beta + \varepsilon$. Similarly if $-2\sup A = \alpha$, then $\forall \varepsilon > 0 $ there exists $x_\varepsilon$ such that $x_\varepsilon > \alpha -\varepsilon$. I'm not sure how to connect the dots here. What can I do to prove this?
 A: I'm assuming the infimum and supremum exist finitely, otherwise that equality won't be true.
Let $\inf (-2A)=a$ and $\sup A=b$. Then $b \geq x$ for all $ x \in A$ this implies $-2b\leq -2x$ for all $x \in A$. $-2b$ is a lower bound of $-2A$ and $a$ is the greatest lower bound of $-2A$, so $a \geq -2b$.
Also $x\leq -\frac{a}{2}$ for all $x \in A$ (as $-2x \geq a)$. So $-\frac{a}{2}$ is an upper bound of $A$. Since $b$ is the least upper bound of $A$ we get $b \leq -\frac{a}{2}$ which implies $-2b \geq a$. Combining we get $a=-2b$.
A: Just do the definitions.  But note, this assumes $A$ is bounded above which, if it is, should be stated.
Let $\alpha = \sup A$.
Prove $-2\alpha$ is a lower bound of $-2A$.
Pf: $\alpha$ is an upper bound of $A$ so for every $a\in A$ we have $a \le \alpha$.  So $-2a \ge -2\alpha$ for all elements of the form $-2a\in -2A$ for some $a\in A$..  As all elements of $-2A$ are of the form $-2a$ for some $a \in A$, we have $-2\alpha \le b$ for all $b \in -2A$.
Prove if $k > -2\alpha$ then $k$ is not a lower bound of $-2A$.
Pf:   If $k > -2 \alpha$ then $-\frac 12k < \alpha=\sup A$.  So $-\frac 12k$ is not an upper bound of $A$.  So there exist an $a\in A$ so that $-\frac 12k < a$.  So $-2a < k$.  But as $a \in A$ we know $-2a \in -2A$. So $k$ is not a lower bound of $-2A$.
So $\inf -2A = -2\alpha$.
That's all.
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To expand on your idea:
If $\inf (-2A) = \beta$ (assuming $\inf(-2A)$ exists) then 1) for any $\epsilon > 0$ there is an $x_\epsilon \in -2A$ so that $\beta \le x_\epsilon < \beta +\epsilon$.  And 2) there are no elements $b \in -2A$ so that $b< \beta$.
As $x_\epsilon \in -2A$ then $x_\epsilon = -2a_\epsilon$ for some $a_\epsilon \in A$.  So $\beta \le x_\epsilon = -2a < \beta +\epsilon$ and $-\frac 12 \beta - \frac 12 \epsilon < a_\epsilon \le -\frac 12 \beta$.
And if $b < \beta$ means $b$ can not be in $-2A$. then $-\frac 12 \beta < -\frac 12 b$ means $-\frac 12b$ can not be in $A$.
So $-\frac 12 \beta$ is an upper bound of $A$ and as for any $\epsilon > 0$ we have that an $a\in A$ must exist where $-\frac 12 \beta -\epsilon < -\frac 12 \beta - \frac 12 \epsilon < a_\epsilon \le -\frac 12 \beta$.
That means $-\frac 12 \beta = \sup A$.
(Of course that assumes $-2A$ is bounded below which we have no reason to assume.)
