Exercise 1.1.6

Question

Show that $\left(\mathbb{R}^{n}, d_{l^{2}}\right)$ in Example 12.1.6 is indeed a metric space.

Prakash Anand · Accepted Answer

(a) We have $$d(x, x)=\left(\sum_{i=1}^{n}\left(x_{i}-x_{i}ight)^{2}ight)^{1 / 2}=\left(\sum_{i=1}^{n} 0ight)^{1 / 2}=0.$$

(b) if $x 
eq y$ then there is some $i$ such that $x_{i} 
eq y_{i}$ and hence we have $\left|x_{i}-y_{i}ight|>0$ and since each term of the following sum is greater than or equal to zero, we get:

$$
d(x, y)=\left(\sum_{i=1}^{n}\left(x_{i}-y_{i}ight)^{2}ight)^{1 / 2}>0
$$

(c) Since $\left(x_{i}-y_{i}ight)^{2}=\left(y_{i}-x_{i}ight)^{2}$ we immediately get that $$d(x, y)=d(y, x).$$

(d) We basically proved this in Exercise 12.1.5. We want to show that $d(x, z) \leq d(x, y)+d(y, z)$. And this is equivalent to showing that:

$$
\left(\sum_{i=1}^{n}\left(x_{i}-z_{i}ight)^{2}ight)^{1 / 2} \leq\left(\sum_{i=1}^{n}\left(x_{i}-y_{i}ight)^{2}ight)^{1 / 2}+\left(\sum_{i=1}^{n}\left(y_{i}-z_{i}^{2}ight)ight)^{1 / 2}
$$

But if we use the final result in \href{./exercise_1-1-5}{Exercise 1.1.5} with $a_{i}=x_{i}-y_{i}$ and $b_{i}=y_{i}-z_{i}$ then we get exactly the above expression.

Exercise 1.1.6

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Exercise 1.1.6

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