TITLE: Continuous bijective way of representing a line on a plane
QUESTION [6 upvotes]: Is there a function $f(a,b)$ which maps ordered pairs to lines in a plane in a continuous, bijective manner? 
Here is the definition I am using for the limit with lines: a sequence of lines $L(1), L(2), \dots$ is said to approach another line $L$ if, for any point $p$ on $L$, the limit as $n\to\infty$ of the distance between $p$ and $L(n)$ is $0$.
If there is no such function, can anyone think of a proof?

REPLY [11 votes]: There is no such bijection.
A line in the plane is almost the same as a plane through the origin in 3-space (by intersecting with the plane at height 1), except there's one plane through the origin that doesn't give you a line (the z=0 plane). So the space of lines in the plane is homeomorphic to $\mathbb{RP}^2$ minus a point: an open mobius strip! So the question is asking if there is a continuous bijection from the open disk $D$ to the open mobius strip $M$. Invariance of domain implies that a continuous bijection between manifolds of the same dimension is a homeomorphism. $D$ and $M$ are not homeomorphic, so there cannot be a continuous bijection between them.