$X=H - \{(0,0)\}$ is contractible where $H =\{(x,y) | y\geq 0\}$ $\mathbf {The \ Problem \ is}:$ Let, $H =\{(x,y) | y\geq 0\}.$ Show $X=H - \{(0,0)\}$ is contractible .
$\mathbf {My \  approach}:$ Let, $p=(0,1).$
Need to show, $id : X \to X$ is homotopic to constant map at $p.$
$X$ isn't convex, so the straight-line homotopy won't work , but intuitively, I can see $X$ can be contracted negotiating $(0,0).$
A small hint on defining the homotopy between them is warmly appreciated .
Thanks in advance .
 A: Hint try to use the usual deformation retract of $\Bbb S^1 \subseteq \Bbb R^2 \setminus \{0\}$, but restricted to the upper half plane.
I can elaborate on request.
A: You don't need convexity.
You only need star convexity.
Fortunately your set $X$ is indeed star convex with respect to the point $p = (0,1)$. In other words, the segment between $p$ and any other point of $X$ is indeed a subset of $X$. Now just do a straight line homotopy using those segments.
A: Convexity is not a topological property. All you have to do is define a homeomorphism  to a convex space, and you can apply the straight-line homotopy to the other space. Remember, a homeomorphism composed with a homotopy is a homotopy (and the inverse of a homeomorphism is a homeomorphism).
Each point in the half plane lies on a line containing the origin, and can be parameterized by the angle of that line and the distance from the origin. Once the origin is removed, the possible distances to the origin constitutes an open interval, and thus is homeomorphic to any other open interval, including finite ones. Thus there is a homeomorphism from the punctured half plane to a rectangular region (for instance, you can take the arctan of the distance to get $(0,\pi)\times(0,\frac{\pi}2)$).
Also, the union of contractible spaces with a contractible intersection is contractible, so if you cut it into contractible spaces, that also proves it's contractible.
