Convergence of random variables Trying to understand the solution given to this homework problem:
Define random variables $X$ and $Y_n$ where $n=1,2\ldots%$ with probability mass functions:
$$
f_X(x)=\begin{cases} 
\frac{1}{2} &\mbox{if } x = -1 \\ 
\frac{1}{2} &\mbox{if } x = 1  \\
0 &\mbox{otherwise}
\end{cases} and\; f_{Y_n}(y)=\begin{cases} 
\frac{1}{2}-\frac{1}{n+1} &\mbox{if } y = -1 \\ 
\frac{1}{2}+\frac{1}{n+1} &\mbox{if } y = 1  \\
0 &\mbox{otherwise}
\end{cases}
$$
Need to show whether $Y_n$ converges to $X$ in probability.
From this I can define the probability space $\Omega=([0,1],U)$ and express the random variables as functions of indicator variables as such:
$X = 1_{\omega > \frac{1}{2}} - 1_{\omega < \frac{1}{2}}$
and
$Y_n = 1_{\omega < \frac{1}{2}+\frac{1}{n+1}} - 1_{\omega > \frac{1}{2}+\frac{1}{n+1}}$
And from the definition of convergence in probability, we need find to show that
$P\{|Y_n-X|>\epsilon\}$ does or does not converge to zero. Which can be written as:
$P\{|1_{\omega < \frac{1}{2}+\frac{1}{n+1}} - 1_{\omega > \frac{1}{2}+\frac{1}{n+1}} - 1_{\omega > \frac{1}{2}} + 1_{\omega < \frac{1}{2}}| > \epsilon \}\;\;(1)$
Now it's easy to see that $\epsilon < 2$ for this to hold, but the solution given states that:
$P\{|Y_n-X|>\epsilon\} = 1 - \frac{1}{n+1} \;\; (2)$
Thus $Y_n$ does not converge in probability to $X$.
My problem is that I don't see the reasoning between (1) and (2). Can anyone shed some insight into intermediate steps/reasoning required to make this step?

You're told that
$$
  P(X=1)=P(X=-1)=1/2 \, ,
$$
and
$$
  P(Y_n=1)=\frac{1}{2} + \frac{1}{n+1}  \;\;\;, \qquad P(Y_n=-1)=\frac{1}{2} - \frac{1}{n+1} \;\;\;,
$$
for $n\geq 1$, and you're asked whether or not $Y_n$ converges to $X$ in probability, which means that
$$
  \lim_{n\to\infty} P(|Y_n-X|\geq \epsilon) = 0 \, , \qquad (*)
$$
for every $\epsilon>0$.
I will assume that $X$ is independent of the $Y_n$'s.
It is not the case that $Y_n$ converges in probability to $X$, because $(*)$ does not hold for every $\epsilon>0$. 
For instance, if we take $\epsilon=1$, then
$$
  P(|Y_n-X|\geq 1)=P(Y_n=1, X=-1) + P(Y_n=-1,X=1) 
$$
$$
  = P(Y_n=1)P(X=-1) + P(Y_n=-1)P(X=1) 
$$
$$
  = \left(\frac{1}{2} + \frac{1}{n+1}\right) \cdot \frac{1}{2} + \left(\frac{1}{2} - \frac{1}{n+1}\right) \cdot \frac{1}{2} = \frac{1}{2} \, ,
$$
for every $n\geq 1$.
