Why is $a$ an accumulation point in $D=(a,b)$?

Question

If an accumulation point $c$ is a point such that $\exists$ a sequence {$x_k$} with $$ \lim_{k\to \infty} x_k = c$$

then how is $a$ an accumulation point for $D=(a,b)$ ? How will there exist a sequence that leads backwards onto $a$?

Answer 1

What about the sequence $x_n=a+\frac1n$?

Answer 2

E.g. $x_k=a+\frac1k$. It tends to $a$.

Answer 3

The sequence $$ a+ \frac {b-a}{n+1} $$ is a sequence in $(a,b)$ which converges to $a$

That makes $a$ an accumulation point of $(a,b)$