Let $X_n\xrightarrow{P} X$ and $Y_n\xrightarrow{P}Y$, then we have $(X_n,Y_n)\xrightarrow{P}(X,Y)$. In process of proof we have the following: $$ \mathbb{P}(\{||(X_n,Y_n)-(X,Y)||\geqslant\epsilon\})\leqslant\mathbb{P}(\{||X_n-X||+||Y_n-Y||\geqslant\epsilon\})\leqslant\mathbb{P}(\{||X_n-X||\geqslant\epsilon/2\})+\mathbb{P}(\{||Y_n-Y||\geqslant\epsilon/2\}) $$ I don't understand the above inequalities. Could someone explain?


$$\|(X_n, Y_n)-(X,Y)\|=\|(X_n-X, Y_n-Y)\|\le \|X_n-X\|+\|Y_n-Y\|$$ is true by triangle inequality.

Hence $$\|(X_n, Y_n) - (X,Y) \| \ge \epsilon \implies \|X_n-X\|+\|Y_n-Y\| \ge \epsilon$$

which explains the first inequality.

Also, $\|X_n - X\| + \|Y_n - Y\| \ge \epsilon$ implies that $\|X_n - X\| \ge \frac{\epsilon}2$ or $\|Y_n - Y\| \ge \frac{\epsilon}2$. Supppose on the contrary that this is not true, then we have $\|X_n - X\| < \frac{\epsilon}2$ and $\|Y_n - Y\| < \frac{\epsilon}2$ and summing them up would give us a contradiction.

Hence, we just have to use a union bound to get the second inequality.

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