If $X$ is Uniform$[0,1]$, does $1-X$ converge in distribution to $X$?

Question

Example 1.1.5 in http://www.datascienceassn.org/sites/default/files/Asymptotic%20Statistics%20-%20Lecture%20Notes.pdf

Here is the example and the question: $X$ ~ $U[0, 1]$, and $X_n = X$ for all n, and $Y_n = X$ for n odd and $Y_n = 1 - X$ for n even. I am confused if $Y_n$ converges to $X_n$ in distribution.

I know it is clearly not, but here is what I deduced by definition of convergence in distribution: $$|F_{Y_n}(t) - F_{X_n}(t)| = |P(Y_n \leq t) - P(X \leq t)| = |P(X > 1-t) - P(X \leq t)| = |1-(1-t)-t| = 0$$

I think there must be something wrong here, but I don't know what it is.

Answer 1

All you need for convergence in distribution is the convergence of the CDFs at every continuity point of the limiting distribution. This clearly holds; in fact, not only do the distributions converge, they are all equal. All the random variables in the sequence are equal in distribution.