Let $a_{n}$ be a sequence which converges to $c$. Then $c$ is a limit point of $a_{n}$ and it is its unique limit point.

Question

Let $(a_{n})_{n=m}^{\infty}$ be a sequence which converges to a real number $c$. Then $c$ is a limit point of $(a_{n})_{n=m}^{\infty}$, and in fact it is the only limit point of $(a_{n})_{n=m}^{\infty}$.

MY ATTEMPT

Let us prove the uniqueness statement first.

Since $a_{n}\to c$, we conclude that $\liminf(a_{n})_{n=m}^{\infty} = c = \limsup(a_{n})_{n=m}^{\infty}$.

But we do also know that any limit point $d$ satisfies $c = \liminf(a_{n})_{n=m}^{\infty} \leq d\leq \limsup(a_{n})_{n=m}^{\infty} = c$. Consequently, $c$ is the unique limit point and we are done.

We may now approach the first part of the statement.

If $a_{n}\to c$, then for every $\varepsilon > 0$, there is a natural number $N\geq m$ such that \begin{align*} n\geq N \Longrightarrow |a_{n} - c| \leq \varepsilon \end{align*}

On the other hand, the definition of limit point says that:

"$c$ is a limit point of $(a_{n})_{n=m}^{\infty}$ iff for every $\varepsilon > 0$ and every $M\geq m$, there is at least one $n\geq M$ such that $|a_{n} - c|\leq\varepsilon$"

But then I get stuck. Could someone please help me to solve it?

Answer 1

Here is another way to show uniqueness.

Suppose that the sequence $\{a_n \}_{n=1}^\infty$ has two limit points $\alpha$ and $\beta$.

Now suppose $\epsilon > 0$ then there exists positive integers $N_1$ and $N_2$ such that $|a_n - \alpha|< \epsilon/2$ for all $n > N_1$ and $|a_n - \beta| < \epsilon/2$ for all $n > N_2$

Hence, for $n > \max\{N_1,N2\}$ we have that $|\alpha - \beta| \leq |\alpha - a_n + a_n - \beta|\leq |\alpha-a_n|+|\beta - a_n| < \epsilon/2 + \epsilon/2 < \epsilon$

Since $\epsilon$ is arbitrary we get that $\alpha = \beta$.

Answer 2

If $ d$ is an other limit point, then we can construct a subsequence $(b_n)=(a_{\phi(n)}) $ of the sequence $ (a_n) $ which converge to $ d$. But $ (a_n)$ converges to $ c, $ thus the subsequence $(b_n)$ converges also to $ c$. we conclude that $ d=c$.

Answer 3

Limit points are unique in any Hausdorff space. But $\Bbb R$ is a metric space. And metric spaces are always Hausdorff.