Show that every Cauchy sequence can have at most one limit point.

Question

Show that every Cauchy sequence can have at most one limit point.

My solution

If a Cauchy sequence $x_{n}$ admits a limit point $L$, then there is a subsequence of $x_{n}$ which converges to $L$. Therefore $x_{n}$ converges to $L$. Since the limit of a sequence is unique in a metric space, the result holds.

Can someone check if is it right? An answer based on $\varepsilon-\delta$ definitions would be nice.

Edit 1

My answer is based on the following result:

Let $(X,d)$ be a metric space and $x_{n}$ is a Cauchy sequence in it. Then, if $x_{n}$ admits a subsequence which converges to $x_{0}$, then the Cauchy sequence is also convergent and $x_{n}\rightarrow x_{0}$.

Edit 2

Since my question has caused some confusion, I am going to explicit what I was looking for.

The proposition "if $L$ is a limit point of the sequence $x_{n}$, then there is a subsequence of $x_{n}$ which converges to $L$" has been proved earlier as well as the other two results involved in the solution.

Maybe I should have asked for an alternative solution involving $\varepsilon-\delta$, but as it has been pointed out it does not make sense since they are already involved in the assumptions used to solve the problem.

Anyway, thank you guys for the feedback.

Answer 1

Any metric space is $T_2$, or Hausdorff. This ensures that sequences have unique limits. A Cauchy sequence is in particular a sequence.

Suppose$x_n\to x$ and $x_n\to y$. Let $U_x,U_y$ be separating neighborhoods for $x$ and $y$. Then $U_x$ contains $x_n$ for all $n\gt N$, for some $N$. But since $U_x$ and $U_y$ are disjoint, it follows that $U_y$ contains only finitely many $x_n$. So $x_n\not\to y$.

The converse is also true. For instance, the cofinite topology on $\Bbb Z$ is a space that is not $T_2$ and limit points aren't unique. For $x_n=n$ converges to every $n$.

Answer 2

A Cauchy sequence is a convergent sequence so it has exactly 1 limit point, not at most 1.

Edit: If the sequence comes from real numbers.

Answer 3

Assume there are two subsequence $ x_{\phi(n)} $ and $ w_{\psi(n)} $ such that

$$x_{\phi(n)}\to L_1$$ and $$x_{\psi(n)}\to L_2$$

Given $ \epsilon>0$,

we know that $$\phi(n)\ge n ; \psi(n)\ge n$$

then there exists $N_1, N_2,N_3 \ge 0$ such that $$n\ge \max(N_1,N_2,N_3) \implies$$

$$|L_1-L_2|=|L_1-x_{\phi(n)} +x_{\phi(n)}-x_{\psi(n)}+x_{\psi(n)}-L_2|\le \frac \epsilon 3+\frac \epsilon 3 +\frac \epsilon 3$$

We conclude that $$L_1=L_2$$

Answer 4

This is standard excercise.

The gyst is: If $\{x_n\}$ has two limit points $L_1$ and $L_2$ then we can take milestones $N_1, N_2$ large enough so that $x_m; m > N_1 $are all arbitrarily close to $L_1$ and $x_n; n > N_2$ are all arbitrarily close to $L_2$ but $L_1$ and $L_2$ are a specific distance apart so the $x_m$ arbitrarily close to $L_1$ are significantly far from the $x_n$ which are arbitrarily close to $L_1$ so the $x_k$ aren't arbitrarily close to each other which contradicts the sequence be Cauchy.

Can you formalize this? Hint:

Triangle inequality says that if $|x_m - L_1| < D_1$ and $|x_n - L_2| < D_2$ then $|L_1 - L_2| = |(L_1 - x_m) + (x_m -x_n) + (x_n - L_2)| \le |x_m-L_1| + |x_m-x_n| +|x_n-L_2| < D_2 + D_1 + |x_m- x_n|$.

But what if $|L_1 - L_2| > D_2 + D_1$. That would be a contradiction, wouldn't it?

Can you formalize that?

Hint: Let $|L_1 - L_2| = K > 0$ and let $\epsilon < \frac K3$.

1) $\{x_n\}$ is cauchy so .... what?

There is an $N_1$ so that if $m,n> N_1$ then $|x_n - x_m| < \epsilon$.

2) $L_1$ is a limit point of $\{x_n\}$ so .... what?

There is an $N_2$ so that if $m>N_2$ then $|L_1 - x_m| < \epsilon$

3) $L_2$ is a limit point of $\{x_m\}$ so ... what?

There is an $N_3$ so that if $n > N_3$ then $|x_n - L_2| < \epsilon$.

So can we formalize: Heres a hint:

Let $N = \max(N_1, N_2, N_3)$.

So what happens if $n,m > N$.

Then $|L_1 -x_m| < \epsilon; |x_m - x_n| < \epsilon; |x_n - L_1| < \epsilon$ so $|L_1 - L_2|\le |L_1 -x_m|+|x_m - x_n|+|x_n - L_1| < 3\epsilon < K = |L_1 -L_2|$.

Answer 5

I stated in a comment below the OP's question that I am a bit confused here. Although the OP also asked for an $\varepsilon \mid \delta$ argument (uh, an $\varepsilon \mid N$), a good idea at this point is to just give a hint for the problem at hand.

PROBLEM; Show that every C̶a̶u̶c̶h̶y̶ $\,$sequence can have at most one limit point.

HINT: Let $a, b$ be any two distinct points in a metric space $X$ and let $l = d(a,b)$. Then

$\quad \{x \in X \mid d(a,x) \lt \frac{l}{2}l \} \; \bigcap \; \{x \in X \mid d(b,x) \lt \frac{l}{2}l \} = \emptyset $