# Proof by induction (finite)

Hi guys I just started metric spaces and was working on the following question. I need to prove by finite induction the following:-

Show that $$x_1,\dots, x_n$$ are n points in a metric space $$(X,d)$$ then

$$d(x_1,x_n) \leq d(x_1,x_2) +\dots+d(x_{n-1},x_n)$$

Now from based on the question by looking at a number of points for example 3 points

$$d(x_1,x_3) \leq d(x_1,x_2)+d(x_2,x_3)$$

therefore by adding a point we would then get the following

$$d(x_1,x_4) \leq d(x_1,x_2)+d(x_2,x_3)+d(x_3,x_4)$$

The following shows my induction proof but I am not familiar with proof by induction and was wondering if I have done it correctly if not can someone guide me:-

$$d(x_1,x_n) \leq d(x_1,x_n)$$ The above is my base case

Initial Hypothesis : Assume the following

$$d(x_1,x_k) \leq d(x_1,x_2)+\dots+d(x_{k-1},x_k)$$

Then the following holds:- $$d(x_1,x_{k+1}) \leq d(x_1,x_2)+\dots+d(x_k,x_{k+1})$$

• Well, how does the last line follow? I can‘t see any justifications in your reasoning. (The inequality is just applying the triangle inequality multiple times which can be formalized via induction.) Nov 14, 2019 at 4:04
• thats what i am trying to do develop the proof via induction but i am not sure how to justify the reasoning and thats what i attempted thus far.
– John
Nov 14, 2019 at 4:11

The base case is: $$d(x_1,x_1)\leq d(x_1,x_1)$$.
The inductive hypothesis is: Assume $$d(x_1,x_k)\leq d(x_1,x_2)+\dots +d(x_{k-1},x_k)$$ for some $$k\geq 1$$.
Then $$d(x_1,x_{k+1})\leq d(x_1,x_k)+d(x_k,x_{k+1})\leq d(x_1,x_2)+\dots +d(x_{k-1},x_k)+d(x_k,x_{k+1})$$