# Somewhat complicated example demonstrating the failure of the Pasting Lemma?

Suppose we let $A_0 = [0,1]$. Now let $A_1$ be the line segment of length $1$ connecting the origin to the Cartesian coordinate $(0,1)$. Define $A_2, A_3, …$ to be similar line segments each of length $1$ connecting the origin to their respective destination points so that all line segments $A_0, A_1, A_2, …$ sweep out the first quadrant in $\mathbb{R}^2$. Now let us denote $F = A_0 \cup A_1 \cup A_2 \cup …$, an infinite union of closed sets. Define the map $f: F \rightarrow \mathbb{R}$ where if the point $p \in A_1$, or $A_2$, or $…$, then $f(p)=0$, and $f((x,0))= x$ on $A_0$. We would like to show that $f|_{A_i}$ is continuous, while $f$ is not. Still not sure about the second case, but for the first case we would have to check that each pre-image restricted to $A_i$ is closed. How could I do that? Guidance would be appreciated!

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You haven’t told us what the destination point of the segments $A_k$ for $k>1$ are. –  Brian M. Scott Sep 20 '12 at 23:08
You are basically creating a "fan" of line segments each of length $1$ like spokes on a wheel, only that were doing this in the first quadrant. The destination points are not predetermined. –  Libertron Sep 20 '12 at 23:19
Actually, I believe the destination points are of the form $(cos \theta, sin \theta)$ where $\theta$ is the angle that each line segment makes with the x-axis. –  Libertron Sep 20 '12 at 23:22
The choice of destination points affects the continuity of the function. –  Brian M. Scott Sep 20 '12 at 23:26
Let $A$ be the set of angles $\theta$ made by the segments other than $A_0$, and view $A$ as a subset of $(0,\pi/2]$. If $0$ is a cluster point of $A$, the function won’t be continuous at any point $\langle x,0\rangle$ with $x>0$; if $0$ is not a cluster point of $A$, the function will be continuous. –  Brian M. Scott Sep 20 '12 at 23:33