TITLE: a continuous bjiective map which is not a homeomorphism
QUESTION [14 upvotes]: Is there a bijective continuous function $f:\mathbb{Q}\rightarrow \mathbb{Q}$ that  not a homeomorphism?
I am not able to prove it or disprove it. The problem that the rationals is not even locally compact.

REPLY [4 votes]: Yes.  Recall that any two countable dense linearly ordered sets without endpoints are order isomorphic, and hence homeomorphic. (See this question for a discussion, or this Wikipedia article for a proof.)
Let $S$ be the subset of $\mathbb{Q}$ consisting of the fractions with odd denominator (in lowest terms), and let $T$ be the complement of $S$.  Then $S$ and $T$ are both countable, dense, and have no endpoints.
Let $U = \mathbb{Q}\cap (-\infty,\sqrt{2})$, and let $V = \mathbb{Q}\cap(\sqrt{2},\infty)$.  Again, $U$ and $V$ are countable and dense, and have no endpoints.
Let $f\colon \mathbb{Q}\to\mathbb{Q}$ be a function that maps $U$ homeomorphically to $S$, and maps $V$ homeomorphically to $T$.  Then $f$ is continuous and bijective, but its inverse is certainly not continuous, so $f$ cannot be a homeomorphism.