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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?

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    $\begingroup$ Do you mean $D=\{(x,y)\in\mathbb R^2~|~xy>0\}$? $\endgroup$ – Hirshy Aug 5 '15 at 10:31
  • $\begingroup$ Yeah your correct. Im engineering student, I know little about mathematical representation, my bad. $\endgroup$ – Salihcyilmaz Aug 5 '15 at 10:35
  • $\begingroup$ Btw, in my opinion, your representation was fully equivalent and it was more obvious which space you meant at first glance. $\endgroup$ – Berci Aug 5 '15 at 10:48
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A domain $D$ is defined as an connected, open subset of a finite-dimensional vector space. So if you have a simple connected domain it must be connected; thus $D$ in your first question is not a domain. And the paths do not need to be simple closed.

But in your second question you still do not have a domain if you include the origin, as $$D=\{(x,y)\in\mathbb R^2~|~xy>0\}\cup\{(0,0)\}$$ is not open.

Edit: why is $$D=\{(x,y)\in\mathbb R^2~|~xy>0\}\cup\{(0,0)\}$$ not a domain?

By including the origin we now have a connected set, but it is not open. For every $\varepsilon>0$ we have that $$(0,-\varepsilon)\in B_{2\varepsilon}(0,0)\quad\text{but}\quad(0,-\varepsilon)\notin D.$$ This means that we can't find $\varepsilon>0$ such that $B_{\varepsilon}(0,0)$ is contained in $D$. Thus $D$ is not open.

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  • $\begingroup$ I think the term 'domain' was rather meant as a 'topological space' in the OP. Nevertheless, your answer is correct in strict terminology. $\endgroup$ – Berci Aug 5 '15 at 10:55
  • $\begingroup$ @Berci, I guess that happens, when you've been dealing with complex analysis for the past few hours... $\endgroup$ – Hirshy Aug 5 '15 at 10:59
  • $\begingroup$ Thats interesting. Then i can not refer to what i say as domain, right? As far as i understand from answer and comment: if there is a domain it is already connected by definition. And connectedness is requirement for simply connectedness. I got it i think. $\endgroup$ – Salihcyilmaz Aug 5 '15 at 11:05
  • $\begingroup$ Can you elaborate on last sentence of your answer? Why we don't have domain, what assumption is not satisfied? $\endgroup$ – Salihcyilmaz Aug 5 '15 at 11:07
  • $\begingroup$ @Salihcyilmaz I put in an explanation why $D$ is not a domain (in the sense of mathematical analysis). $\endgroup$ – Hirshy Aug 5 '15 at 11:17
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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.

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