Cfrgtkky

True or False:
If E is a subset of a metric space X, then any sequence of points of E that is a Cauchy sequence in X is a Cauchy sequence in E.

I'm having difficulty understanding this language. What does it mean for a sequence to be Cauchy "in" X?

Since $X$ is a metric space there is a distance $d:Xtimes Xto [0,+infty)$. Then ${x_n}_nsubset X$ is a Cauchy sequence in $X$ (or with respect to this distance $d$) if $$forallepsilon>0,; exists N;:; forall n,m>N, d(x_n,x_m)<epsilon.$$
If $Esubset X$ and ${x_n}_nsubset E$ then we may say that if ${x_n}_n$ is a Cauchy sequence in $X$ then it also a Cauchy sequence in $E$. $E$ and $X$ share the same distance if nothing else has been specified.

The definition of Cauchy sequence is$$forall epsilon>0 exists Nin mathbb Ntext{ such that } d(a_n,a_m)<epsilon forall n,m>N$$
Now, the existence of this N is dependent on the metric space(similar to the fact that limit points of a set may or may not be in a particular metric space).

Coming back to the question, yes, the given statement is true.

$because$We know that the sequence is in E, therefore, all of the terms of the sequence is in E. Hence, all the terms after the stage N are also in E and therefore the property satisfies.

Hope it helps:)

The same sequence may be Cauchy in one metric space but not Cauchy in another. For example, let $M =mathbb{Q}$ with the normal distance function, $d(x, y) = |x-y|$, and let $N = mathbb{Q}$ with $d(x, y) = 1$ if $x not = y$ and $d(x, x) = 0$ (this is called the discrete metric).

Now notice that $M$ and $N$ have the same set, but they are different metric spaces since a metric space is the set together with the distance function.

Now consider $x_n = dfrac 1n$. In $M$, this sequence is convergent and therefore Cauchy (the terms bunch up near $0$ and thus get closer and closer to each other).

Now since the sequence never repeats terms, in $N$, every member of the sequence is at distance $1$ from every other member, so the terms do not get closer and closer.

Thus $x_n$ is Cauchy in $M$ but not Cauchy in $N$.

Now your exercise is one of those "so easy that's it's hard". There isn't really anything to do to solve it, but you do have to understand exactly what is going on to write a correct answer.