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Consider the subset of $\mathbb{R}^2$: $$T=\{(x,y):x\notin \mathbb{Q} \text{ or } y \notin \mathbb{Q}\}$$

Is $T$ open, closed, compact, connected (under the standard Euclidean topology)?

First, $cl(T)=\mathbb{R}^2$ and hence $T$ is not closed.

Then since $\mathbb{Q}$ is dense in $\mathbb{R}$, it's not possible to find an open neighborhood $U$ of $(x,y)\in T$ such that $U\subset T$. Hence $T$ is not open.

Consider the open cover $\{B(x,1)\cap T: x\notin \mathbb{Q}\}$, which is the union of open balls centered at irrational $x$ with radius $1$. Then $T$ is not compact.

Is $T$ connected? $T$ is the $xy-$plane after deleting all rational points. Hence I think $T$ is more than connected: $T$ is path-connected.

SHBaoS
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You are right, and in fact an even smaller subset is connected, namely the set of points with two irrational coordinates. For seeing this, consider $x,y$ such points. Then, the set of circles passing through $x$ and $y$ is uncountable : in particular, there is one circle which does not contains any rational coordinate (else this would define an injection $\mathbb R \to A$ with $A \subset \mathbb Q$) and so your space is path-connected.