Show convergence in probability of the reciprocal I have the following problem-
If $Y_n$ converges in probability to $Y$ then show that $\frac{1}{Y_n}$ converges in probability to $\frac{1}{Y}$. Also $P(Y_n=0) = 0$ for all $n$ and $P(Y = 0) = 0$
My attempt - 
Since $Y_n$ converges in probability to $Y$ then we know that $Y_n$ is bounded in probability
and $Y$ is also bounded in probability , so I can get $M$ s.t.
$$P(|Y_nY| > M) < \eta$$
then $P(|\frac{1}{Y_n} - \frac{1}{Y}| > \epsilon) = P(\frac{|Y_n-Y|}{|Y_nY|} > \epsilon) \leq  P(\frac{|Y_n-Y|}{|Y_nY|}>\epsilon,|Y_nY| > M) +P(\frac{|Y_n-Y|}{|Y_nY|}>\epsilon,|Y_nY| \leq M) \leq P(|Y_nY| > M) + P(\frac{|Y_n-Y|}{|Y_nY|}>\epsilon,|Y_nY| \leq M)  $
I don't know how to go ahead from here,  basically the problem is the second term in the last expression,  if I can somehow manipulate that to use the convergence of $Y_n$ I'd be done.
So can someone please provide me with the solution to this problem
Edit: I have made the correction and now it says bounded in probability not just bounded.
 A: For any $\epsilon>0,\delta>0$, we have
$$P\left(\left|\frac1{Y_n}-\frac1Y\right|>\epsilon\right)=P\left(\left|\frac{Y_n-Y}{Y_nY}\right|>\epsilon,|Y_nY|\leq \delta\right)+P\left(\left|\frac{Y_n-Y}{Y_nY}\right|>\epsilon,|Y_nY|> \delta\right)\leq P(|Y_nY|\leq\delta)+P(|Y_n-Y|>\delta\epsilon).$$
By the convergence in probability, $\lim_{n\to\infty}P(|Y_n-Y|>\delta\epsilon)\to0$. All we have to do is prove that $\limsup_{n\to\infty}P(|Y_nY|\leq\delta)\to0$ as $\delta\to 0$.
For any $\xi>0$, we have
$$P(|Y_nY|\leq\delta)=P(|Y_nY|\leq\delta,|Y_n-Y|\leq \xi)+P(|Y_nY|\leq\delta,|Y_n-Y|>\xi)\leq P(|Y_nY|\leq\delta,|Y_n-Y|\leq \xi)+P(|Y_n-Y|>\xi).$$
If $|Y_nY|\leq\delta$ and $|Y_n-Y|\leq \xi$, then $|Y_n|\geq |Y|-\xi$, so $$\delta\geq |Y_n||Y|\geq |Y|(|Y|-\xi)=\left(|Y|-\frac{\xi}2\right)^2-\left(\frac\xi2\right)^2,$$
and $\frac\xi2-\sqrt{\delta+\frac{\xi^2}4}\leq |Y|\leq\frac\xi2+\sqrt{\delta+\frac{\xi^2}4}.$ Hence
$$P(|Y_nY|\leq\delta,|Y_n-Y|\leq \xi)\leq P\left(\frac\xi2-\sqrt{\delta+\frac{\xi^2}4}\leq |Y|\leq\frac\xi2+\sqrt{\delta+\frac{\xi^2}4}\right)\leq P\left(|Y|\leq\frac\xi2+\sqrt{\delta+\frac{\xi^2}4}\right).$$
Therefore
$$\limsup_{n\to\infty}P(|Y_nY|\leq\delta)\leq P\left(|Y|\leq\frac\xi2+\sqrt{\delta+\frac{\xi^2}4}\right)$$
for all $\xi>0$, which means
$$\limsup_{n\to\infty}P(|Y_nY|\leq\delta)\leq P(|Y|\leq\sqrt\delta)$$
which converges to $0$ as $\delta\to0$ since $P(Y=0)=0$. The proof is complete now.
