I have the following problem:

Prove that the euclidean distance in $\mathbf R^n$ is uniformly continuous by two variables simultaneously, i.e. it is a continuous function of the point $(P, Q) \in \mathbf R^{2n}$.

My understanding of the problem is the following:

I have to prove that $\forall \epsilon >0 \exists \delta > 0$ s.t. for every two pairs of points $(P_1, Q_1)$ and $(P_2, Q_2)$ we have that $d((P_1, Q_1), (P_2, Q_2))<\delta(\epsilon) \implies |d(P_1, Q_1)-d(P_2, Q_2)| < \epsilon$.

My attempt:

Let $d((P_1, Q_1), (P_2, Q_2))< \delta$.

Then from $|x-y| \le |x|+|y|$ we have

$$ |d(P_1, Q_1)-d(P_2, Q_2)| \le |d(P_1, Q_1)| + |d(P_2, Q_2)|=\\=d(P_1, Q_1)+d(P_2, Q_2)=\\ \sqrt{\sum_{i=1}^n(x'_i-y'_i)^2}+\sqrt{\sum_{i=1}^n(x''_i-y''_i)^2} $$

That last part I'm trying to express in terms of $\sqrt{\sum_{i=1}^n(x'_i-y'_i)^2+\sum_{i=1}^n(x''_i-y''_i)^2}$ which I believe is the distance between $(P_1, Q_1)$ and $(P_2, Q_2)$

Thanks in advance.


This is true in any metric space. In fact you universally have $$|d(x,y)-d(x_0,y_0)|\leq d(x,x_0)+d(y,y_0)\ .\tag{1}$$ Proof. By the triangle inequality $$d(x,y)\leq d(x,x_0)+d(x_0,y_0)+d(y_0,y)\ ,$$ hence $$d(x,y)-d(x_0,y_0)\leq d(x,x_0)+d(y,y_0)\ .$$ Interchanging $(x,y)$ and $(x_0,y_0)$ gives also $$d(x_0,y_0)-d(x,y)\leq d(x,x_0)+d(y,y_0)\ ,$$ so that we obtain $(1)$.

If you compute the distance of $(x,y)$ from $(x_0,y_0)$ using Pythagoras' theorem you can use $d(x,x_0)\leq d\bigl((x,y),(x_0,y_0)\bigr)$, and similarly for $d(y,y_0)$. It follows that $(1)$ can then be replaced by $$|d(x,y)-d(x_0,y_0)|\leq 2 d\bigl((x,y),(x_0,y_0)\bigr)\ .$$

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  • $\begingroup$ Maybe it's a stupid question, but how to express $d((x, y), (x_0, y_0))$ in terms of $d(x,x0)+d(y,y0)$? $\endgroup$ – Nikola Apr 5 '18 at 8:27

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