Is $\limsup x_n = a \iff \limsup |x_n - a |= 0$ true?

Question

We know that $\lim x_n = a$ is equivalent to $\lim |x_n-a|=0$.

I am trying to show that $$\limsup x_n = a \iff \limsup |x_n - a |= 0$$

Define $s_n := \sup_{m \geq n} x_m$. By definition, $\limsup x_n = \lim s_n$. So $$\limsup x_n = a \iff \lim s_n = a \iff \lim |s_n - a|=0$$ Note that: $s_n - a = \sup_{m \geq n} x_m - a = \sup_{m \geq n} \{x_m - a\}=: \bar s_n$. So: $$\limsup x_n = a \iff \lim |\bar s_n|=0 $$

Moreover, defining $\tilde s_n := \sup_{m \geq n} |x_m - a|$ $$\limsup |x_n - a |= 0 \iff \lim \tilde s_n = 0$$

It remains to show that $$\lim |\bar s_n|=0 \iff \lim \tilde s_n = 0 $$

But I think that it is not true. Do you have a counterexample? or... is this true?

Answer 1

False. Counter-example: $x_n=(-1)^{n}, a=1$.

Answer 2

For a counter example, see the other answer. I am writting this answer to give just some explanations that does not fit into a comment in my opinion.

$\lim \sup |x_n-a|=0$ is in fact equivalent to $\lim x_n = a$.

Forward implication is given by: $ 0 \le |x_n-a| \le \sup_{m \ge n} |x_m-a| \to 0$, hence convergence by the squeeze theorem. (And the converse is trivial from the definition of $\lim \sup$).

Answer 3

As geetha290km said in their answer, the claim is false and any piece-wise function is a proof of this. However, I have a feel that the statement without the absolute value bars (below) holds: $$\limsup x_n=a \iff \limsup (x_n-a)=0$$ Because for other members $x_{n^*}$ (e.g. for $n$-s, for which $x_{n^*}=\liminf x_n=b$, it holds: $$x_{n^*}-a=b-a<0=\limsup (x_n-a)$$