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Let $X=\{(x,y)\in \mathbb{R^2} \text{ s.t. } x \in [0,1], y=1/n, n \in \mathbb{N}\}$

And let $\sim$ the following equivalence relationship, $x\sim y \iff$ $x$ and $y$ lie in the same coneccted component. Show that $X/{\sim}$ is not Hausdorff. First I notice how is the set, not coneccted. Then I have a couple of questions:

1) What does it mean to lie in the same coneccted component, $x$ belongs to the union of all coneccted set that contains $y$?

2) This coneccted set is coneccted related to $X$ or the whole space? For example what is the $[(0,0)] \text{?}$

3) I know that I have to find two equivalence classes such that for every two open sets have at least one point in common. But the open sets in the quotient are open sets in the space. What is the equivalence class of an open set?

Thanks and sorry if I did a mistake in the language ;)

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    Notice that $\{\text{curly braces}\}$ belong inside the MathJax code, thus $$\{(x,y)\}$$ or $\displaystyle \left\{ \sum_{i=1}^\infty 1 \right\},$ and a subtler point: In the expression $X/\sim$ there's more space than there should be after the slash because usually that last symbol is a binary operator, as in $x\sim y.$ So code it as X/{\sim} and what you see is $X/{\sim}$ (because there's nothing where $x$ and $y$ were in $x\sim y$). $\qquad$ – 2017-01-21
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    As written, if $X$ gets the subspace topology and $X/{\sim}$ the quotient topology, then $X/{\sim}$ is Hausdorff. Is it perhaps $$X = \{(x,y) \in \mathbb{R}^2 : x \in [0,1], y = 1/n \text{ for some } n \in \mathbb{N}\} \cup \{(0,0),(1,0)\}\,?$$ – 2017-01-21
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    Oh yes sorry I forgot to put these two points. – 2017-01-22

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