I have to prove that a constant sequence and an eventually cinstant sequence is always convergent.

I tried to do this as follows :

I considered (xn) in X to be a constant sequence such that (xn) = x,x,x.... for all n in N.

Then i considered an open ball centered at x and with radius 1/2 i.e. b[x;1/2).

Since this ball will definitely contain a tail of the sequence (xn) therefore (xn) is convergent.

Is this argument correct.Any arbitrary value of "r" should work, right ?

Also can the convergence of eventually constant sequences be proved on the same lines? How do i decide what value of "r" should be chosen for this case ?


Your argument is not quite correct. You should start with an arbitrary value of $r>0$, at which point your conclusion follows. (Since $r>0$ was arbitrary, then every open ball about $x$ contains a tail of the sequence, so we have convergence.)

As for the eventually constant sequence, you know that there is some $x$ and some $N$ such that $x_n=x$ for all $n\ge N$. The same sort of argument works here, with an arbitrary $r>0$.

To see why we need $r>0$ to be arbitrary, consider the sequence of real numbers $x_n=(-1)^n$. Now, $B[0;2)$ contains every point of the sequence, but $B[0;1)$ contains none of the points. In fact, for every $x\in\Bbb R,$ there is an $R>0$ such that $B[x;R)$ contains every point of the sequence, but there is also some $r>0$ such that $B[x;r)$ does not contain a tail of the sequence, so the sequence does not converge.

  • $\begingroup$ So is it sufficient to say that the open ball b[x;r) where r is arbitrary will always contain a tail of the sequence (xn) and hence a constant sequence is always convergent. $\endgroup$
    – johny
    Oct 1 '13 at 13:19
  • $\begingroup$ Absolutely fine. $\endgroup$ Oct 1 '13 at 13:22

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