Why $f(x^k) - f(x^{k+1}) \rightarrow 0$ for monotonically non-increasing sequence $\{f(x^k)\}$?

Question

Given $\{f(x^k)\}$ a monotonically non-increasing sequence $\{f(x^k)\}$ converges to a finite value or diverges to $-\infty$. Since $f$ is continuous, $f(\bar{x})$ is a limit point of $\{f(x^k)\}$, so it follows that the entire sequence $\{f(x^k)\}$ converges to $f(\bar{x})$ and $f(x^k) - f(x^{k+1}) \rightarrow 0$.

My questions are:

1. Why $f(\bar{x})$ is a limit point of $\{f(x^k)\}$?
2. Why the entire sequence $\{f(x^k)\}$ converges to $f(\bar{x})$?
3. Why $f(x^k) - f(x^{k+1}) \rightarrow 0$?
4. Given that $\{x^k\}$ converges to a stationary point $\bar{x}$ why $f(x^k) - f(x^{k+1}) \rightarrow 0$?

Thanks!

Answer 1

1) $(x_k)$ has a subsequence converging to $\bar x$. Since $f$ is continuous, the same subsequence now taken from $(f(x_k))$ converges to $f(\bar x)$.

2) This is a property of monotonic sequences.

3) If $f(x_k)\to f(\bar x)$, then also $f(x_{k+1})\to f(\bar x)$.

4) Follows from (1)-(3). Has nothing to do with stationarity of $\bar x$.