Convergence in Probability of maximum of r.v. Suppose that $\{X_n^{(1)}\}, \ldots, \{X_n^{(k)}\}$ are sequences of random variables that $X_n^{(i)}\rightarrow_p 0$ as $n\rightarrow \infty$ for each $i=1,...,k$.
I have to show that  $ \max_{1\leq i\leq k}|X_n^{(i)}|\rightarrow_p 0$ (assuming $k$ fixed), and when $k$ increases with n, i.e., $k=k(n)$, give a counter-example.
I'm not sure how to write/formalize my answer, or if it's even correct. 
For the first part I thought of proving that the maximum is a continuous function, and so by the continuous map theorem, I would obtain what I want. However, because I cannot assume that the each sequence is independent of the other, I have no idea how to prove continuity.
So I decided to reason the following way:
$$ 0\leq P\left\{\max_{1\leq i\leq k}|X_n^{(i)}|\leq \varepsilon\right\}\leq P\left\{ \bigcup^k_{i=1}\left\{|X_n^{(i)}|\leq \varepsilon\right\}\right\}\leq \sum^k_{i=1}P\left\{|X_n^{(i)}|\leq \epsilon\right\}\xrightarrow[n\rightarrow \infty]{} 0 $$
This reasoning is correct, no?
For the second part, I was thinking of something like defining every sequence as $\{X_n^{(i)}=\frac i n \}$ and $k=n$. This way we would get convergence in probability to zero for every sequence, and because $\max_{1\leq i\leq n}|X_n^{(i)}|=1$, the maximum would not converge in probability to zero. This reasoning seems intuitively correct. However I do not know how to formalize it.
Any help would be appreciated.
 A: For the first part (with $k$ fixed):
Since $X_n^{(i)}\rightarrow_p 0$ for $i=1,2,\dots,k$ we have
$$P\left\{|X_n^{(i)}|> \epsilon\right\}\xrightarrow[n\rightarrow \infty]{} 0$$
For any $\delta >0$, there exists $N_i$ such that when $n \geq N_i$
$$P\left\{|X_n^{(i)}|> \epsilon\right\}< \delta/k.$$
Hence if $n \geq \max(N_1, N_2, \ldots, N_k)$ then
$$0\leq P\left\{\max_{1\leq i\leq k}|X_n^{(i)}|> \varepsilon\right\}= P\left\{ \bigcup^k_{i=1}\left\{|X_n^{(i)}|> \varepsilon\right\}\right\}\leq \sum_{i=1}^{k}P\left\{|X_n^{(i)}|> \epsilon\right\}< \delta.$$
For the second part:
Let $X_n^{(i)}$ be independent binary random variables for $i = 1,2, \ldots, n$ with
$$P(X_n^{(i)}=1) = \frac1{n},\\P(X_n^{(i)}=0) = 1-\frac1{n}.$$
For fixed $i$ we have $X_n^{(i)}\rightarrow_p 0$ and for $\epsilon < 1$
$$P\left\{\max_{1\leq i\leq n}|X_n^{(i)}|> \varepsilon\right\}= P\left\{ \bigcup^n_{i=1}\left\{|X_n^{(i)}|> \varepsilon\right\}\right\}=\sum_{i=1}^{n}{n \choose i}(-1)^{i-1}n^{-i}\\=1-\sum_{i=0}^{n}{n \choose i}(-1)^{i}n^{-i}= 1 - \left(1-\frac1{n}\right)^n.$$
I used the principle of inclusion and exclusion above, but the result is more obvious for independent random variables as
$$P\left\{\max_{1\leq i\leq n}|X_n^{(i)}|> \varepsilon\right\}= P\left\{ \bigcup^n_{i=1}\left\{|X_n^{(i)}|> \varepsilon\right\}\right\}\\= 1- P\left\{ \bigcap^n_{i=1}\left\{|X_n^{(i)}|> \varepsilon\right\}^C\right\}=1 - \left(1-\frac1{n}\right)^n.$$
Hence, 
$$\lim_{n \rightarrow \infty}P\left\{\max_{1\leq i\leq n}|X_n^{(i)}|> \varepsilon\right\}=1-e^{-1}>0$$
