Why $\mathbb Q$ is not locally compact, connected, or path connected? [closed]

for every $a,b \in \mathbb{R}, \space a \neq b$, the open subset $(a,b)_\mathbb{Q} \equiv (a,b) \cap \mathbb{Q}$ can be partitioned given some irrational $c$ for which $a<c<b$ as : $$(a,b)_\mathbb{Q}=(a,c)_\mathbb{Q} \cup (c,b)_\mathbb{Q}.$$ So, there are no connected sets in $\mathbb{Q}$, and therefore it is not locally/connected.

As for local/compactness:
Every open set of the above form can be partitioned to infinite open set elements (Serving as an infinite cover with no finite subcover), and therefore every (non empty) subset of $\mathbb{Q}$ containing an open set is not compact. Given that all neighborhoods contain an open set, it follows that all neighborhoods contained in $\mathbb{Q}$ are not compact. The conclusion is that $\mathbb{Q}$ is not locally/compact.

Also, refer to Arthur Fischer's answer.

To show that $\mathbb{Q}$ is not locally-compact, you could make use of the statement from this question:

A subset $M$ of a locally-compact (Hausdorff) space $X$ is itself locally compact iff it is of the form $U \cap F$ for some open $U \subseteq X$ and some closed $F \subseteq X$.¹

Now suppose that $U \subseteq \mathbb{R}$ is open and $F \subseteq \mathbb{R}$ is closed such that $\mathbb{Q} = U \cap F$. Since $\mathbb{Q}$ is dense in $\mathbb{R}$ and $\mathbb{Q} \subseteq F$, it follows that $F = \mathbb{R}$. But then $U = \mathbb{Q}$, and we know that $\mathbb{Q}$ is not open in $\mathbb{R}$! Therefore $\mathbb{Q}$ is not locally-compact.

(I can't think of a proof of the non-local-connectedness of $\mathbb{Q}$ which substantially differs from Dror's. I will point out that Dror's answer shows that $\mathbb{Q}$ is totally disconnected: the only connected subsets are the singletons. And no non-discrete totally disconnected space is locally-connected.)

¹Actually, you don't need to assume that the original space is locally-compact; Hausdorff suffices.

Another argument for the local compactness : it's not locally compact, because it's not a Baire space, and all locally compact Hausdorff spaces are Baire spaces.

The components of $\mathbb{Q}$ are singletons (because between every two rationals lies an irrational that separates them, essentially), which kills connectedness, and every open subset of $\mathbb{Q}$ is homeomorphic to $\mathbb{Q}$, and so this also holds for all open subsets. So the space is not locally connected (which is equivalent to every component of an open subset is open), because $\mathbb{Q}$ has no isolated points.