<p>In Rⁿ, define</p> <p>d\'(<strong>x,y</strong>) = |x₁ -

Question

<p>In Rⁿ, define</p>
<p>d'(<strong>x,y</strong>) = |x₁ - y₁|+...+|x_n - y_n|.</p>
<p>Show that d' is a metric that induces the usual toplogy of Rⁿ.</p>
<p>Can someone please explain this starting with what needs to be shown? I mean so that I can follow it. &#160;I don't understand the answer that is posted on cramster b/c I can't follow the logic. Please include intermediate steps.</p>

Explanation / Answer

We are firstly trying to prove that some func d' is a metric on Rn
That involves 4 easy steps
d'(x,y) = 0 iff x =y here this is easy as |xi-yi|=0 iff xi=yi;
Another is d'(x,y) >= 0 for all x,y , true again
d'(x,y) + d'(y,z) >= d'(x,z)
This is true since |xi-yi| + |yi - zi| >= |xi - zi| for all i
So summing over it we have the triangle inequality

So this finishes the proof that d' is a metric on Rn

Given any metric d we can induce a topology on X using d
consider d(x,y)< k it is a ball of size k(called a ball, its shape need not be ball like, cause in the problem its a n dimensional cube each of its side k)

Now consider the metric in the problem
Let d be the usual euclidian metric = d(x,y) = sqrt (sum over i (xi-yi)^2)
and d' is the given metric

Let B(d,x,k) = {set of y st d(x,y)<k}
Consider lemma 20.2 : on page 138 :
We say one topology(d') is finer that other(d) if for each x,k there exists l st
B(d',x,l) C B(d,x,k)

Now i will give a proof for the problem which closely resembles the proof on page 139
Now d'(x,y)/sqrt(n) <= d(x,y) <= sqrt(n) * d'(x,y)
For the first part is called the rms am inequality as sum |xi-yi|/n <= sqrt((1/n)sum |xi-yi|^2)
And the second part is : d'(x,y) >= max|xi-yi|=M, but nM^2 >= sum |xi-yi|^2
So d(x,y) <= sqrt(n) * d'(x,y)
Hence we have shown that topology generated by d' is finer that that gen. by d in first inequality
and that topology generated by d is finer that d' in second (refer to proof on 139 thm 20.3)
So the topologies are equal

I will be online for some more time, ask if you have any doubts

Related Questions

Navigate

• Browse (All)
• Browse #
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.