Is the space $Y=\{(a,b)\in\mathbb{R}^2:a+b\in \mathbb{Z}\}$ a connected space? 
Is the space $Y=\{(a,b)\in\mathbb{R}^2:a+b\in \mathbb{Z}\}$ a connected space?

I believe the space is connected because I think the relation that $(a,b)\sim(x,y)$ if and only if $a+b \text{ and } x+y$ are not in $\mathbb{Z}$, is an equivalence relation.
So the quotient map would preserve connectedness from $\mathbb{R}^2$, But I would need to show that $\mathbb{R}^2/ \sim$ is homeomorphic to $Y$.
 A: Your idea doesn't work in this specific case, as I'll show below. But it is a good idea. Hope you'll have a chance to make it useful elsewhere.

We prove by contradiction that $Y$ is not connected.
Let $f:Y\rightarrow\mathbb{R}$ be the function that sends $(x,y)$ to $(x+y)$. $f$ is continuous, hence if $Y$ is connected, $f(Y)=\mathbb{Z}$ would be connected, which is a contradiction.
In fact $Y$ has countably many connected components, namely the lines $x+y=n$.


The (set) quotient $A$ of $\mathbb{R}^2$ by your relation is the set $Y\sqcup\{*\}$, where $*$ is the equivalence class of all points not contained in $Y$. A non empty open set of $A$ necessarily contains $*$ (for otherwise its preimage is contained in $Y$ of $\mathbb{R}^2$, which cannot be open, except for empty set). Hence $A$ is indeed connected. $Y$, as a subspace of $A$, is also connected (there aren't many open sets to divide $Y$, you killed them all with your enormous equivalence class).

By connectedness, now we know that neither the quotient $A$ nor its subspace $Y$ is homeomorphic to $Y$.
A: Consider $f:\mathbb{R}^2\rightarrow\mathbb{R}$ defined by $f(a+b)=a+b$ it is continuous, $Y=\cup_{n\in\mathbb{Z}}U_n$ where $U_n=f^{-1}(n)$, $U_n$ are closed non empty and disjoint and $Y$ is not connected.
