Algebraic sum of open sets For a non-empty set $A \subseteq \mathbb{C}$ and a $t \in \mathbb{C}$ we define $A+t= \{ a+t \vert a \in A \}$. Show that, if $A$ is open, then $A+t$ is open for all $t \in \mathbb{C}$.
My attempt: Take a $a \in \mathbb{C}$. As $A$ is open, there exists a $\delta >0$ so that $]a- \delta ,a+ \delta [ \subseteq A$. Now if we want to prove that $A+t$ is open, there has to exist a $\delta_{1} > 0$ so that $] a+t- \delta_{1} ,a+t+ \delta_{1} [ \subseteq A +t$. Now it is clear to me that $\delta_{1} = \delta$, but I am kind of lost on how to get there.
 A: Let $A \subseteq \mathbb{C}$ be open and non-empty and $t \in \mathbb{C}$. 
Now let $\alpha \in A +t$ be given, we must find a $\delta_1 >0$ such that $$B(\alpha,\delta_1)  \, \, \subseteq A+t$$As $\alpha \in A+t$ we can write $\alpha = b+t$ for some $b \in A$. Since $A$ is open we know there exists $\delta >0$ such that $$B(b,\delta) \subseteq A $$ Set $\delta_1 = \delta$ and let $x \in  B(\alpha,\delta)=B(b+t,\delta)$.   Then $$x-t \in B(b+t,\delta)-t=B(b,\delta) \subseteq A$$
Thus $(x-t)+t = x \in A+t$ and it follows that $B(\alpha,\delta_1)  \, \, \subseteq A+t$ implying $A+t$ is open.
A: We can just view this as applying the map
$$
f(x) = x+t
$$
and asking is $f(A)$ open when $A$ is. 
Now $f$ has the inverse
$$
g(x) = x-t.
$$
The inverse image of an open set under a continuous mapping is continuous so 
$$
g^{-1}(A)
$$
is open and we are done.
How do we prove: The inverse image of an open set under a continuous mapping is continuous . This is actually a definition of continuity... 
A: (I). One commonly-used way to show that a set $S$ is open is to show that for each point $p\in S$ there exists an  open set $U(p)$ with $p\in U(p)\subset S....$ 
....Because $S\supset \cup_{p\in S} U(p)\supset \cup_{p\in S}\{p\}=S,$ so $S=\cup_{p\in S}U(p)$  is the union of the family $\{U(p)|p\in S\}$ of open sets, so $S$ is open.
(II). Let $A$ be an open subset of $\mathbb C.$ Let $t\in \mathbb C.$ For each $p\in A+t$ let $p'=p-t.$  Then  $ p'\in A....$ 
....And take $r_p>0$ such that $B(p',r_p)=\{z\in C:r_p>|z-p'|\} \subset A.$ Let $U(p)=B(p',r_p)+t.$ Then $$p\in U(p)\subset A+t.$$ 
Now for any $z\in \mathbb C$ we have $|z-p|<r_p\iff$ $ |(z-t)-(p-t)|<r\iff$ $ |(z-t)-p'|<r_p\iff$ $ z-t \in B(p,r_p)\iff$ $ z=(z-t)+t\in B(p,r_p)+t=U(p).$ Hence   $$U(p)=\{z\in\mathbb C:r_p>|z-p|\} \quad \text {is open.}$$ Therefore  $A+t=\cup_{p\in A}U(p)$ is open.
