does simply connectedness require connectedness? My question consists of two parts.
$1)$
suppose domain $D=\{(x,y)\in\mathbb R^2~|~xy>0\}$ is given. Now that is first quadrant and third quadrant with exclusion of $x$ and $y$ axis. We can easily see that $D$ is not connected, since there is a discontinuity at origin. But every closed curve we can construct in domain contains interior of it (or formally, they can shrunk to a point). So do we call it simply connected, or do we also need connectedness to say $D$ is simply connected?
$2)$
Now for second part, let origin also included in domain so that $D$ is connected. Let us construct closed curve which goes through origin. Now this curve also satisfies assumption given above. But it is not simple closed curve. Does this affects simply connectedness? Do we need simple closed curves for simply connectedness?

According to the Wikipedia, that definition of simply connectedness excludes all the non path connected spaces. 
The injectivity of the closed curves is not mentioned and not required. 
Your second example thus becomes simply connected.
