In $\mathbb{R}^n$, suppose $A$ is an open set and $B$ is any set. Show that the set $A + B = \{a + b : a ∈ A, b ∈ B\}$ is open.
For the solution i'm stuck at this point: In $\mathbb{R}^n$, what is $B(p, ε) + \{b\}?$
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2$B(p+b,\epsilon).$ – Thomas Andrews Sep 25 '22 at 01:23
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Tacking onto @ThomasAndrews's comment: To see it, that is exactly $B(p, \varepsilon) + {b} = {a+b : a \in B(p, \varepsilon)}$. I.e., you've translated the ball over by $b$. – Brian Tung Sep 25 '22 at 01:26
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@BrianTung so its just an open ball with center $p+b$? – Hitman Sep 25 '22 at 01:46
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As is said in the comments, it is just the open ball around $b+p$ and radius $\varepsilon$: $$\begin{align*} B(p,\varepsilon) + \{b\} &= \{a+b : a \in B(p,\varepsilon)\} \\ &= \{x \in \Bbb R^n : \exists a \in B(p,\varepsilon),\, x=a+b\} \\ &= \{x \in \Bbb R^n : \exists a \in B(p,\varepsilon),\, x-b=a\} \\ &= \{x \in \Bbb R^n : x-b \in B(p,\varepsilon)\} \\ &= \{x \in \Bbb R^n : \|(x-b)-p\|<\varepsilon\} \\ &= \{x \in \Bbb R^n : \|x-(b+p)\|<\varepsilon\} \\ &= B(b+p,\varepsilon). \end{align*}$$
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