Define a sequence $\{y_n\}$ by $y_{2n-1}=x_n$ and $y_{2n}=x$. Which of the following are true?

Question

Let $(X,d)$ be a metric space and let $x_n$ be a sequence in $X$. Let $x\in X$. Define a sequence $\{y_n\}$ by $$y_{2n-1}=x_n,y_{2n}=x,n\in \mathbb N$$ Which of the following statements are correct?

(A) If $x_n\to x$ as $n\to \infty$, Then the sequence $\{y_n\}$ is Cauchy.

(B) If the sequence $\{y_n\}$ is cauchy, then $x_n\to x$ as $n\to \infty$.

(C) Let $f:X\to X$ be a mapping that maps Cauchy sequences to Cauchy sequences. Then $f$ is continuous.

My attempt

(A) If $x_n\to x$. Hence, $y_n \to x$. Every convergent sequence is cauchy.

(B) Since Given space is not complete, We know that $\{y_n\}$ is Cauchy $\implies$ $\{x_n\}$ is Cauchy. But need not converges. I am not able to cook counter examples.

(C) We want to prove $f: X\to X$ is continuous. $x,y\in X$. Let us try to prove by contradiction. Suppose $f$ is not continuous. So, there is $x\in X$,there is an $\epsilon_o \in \mathbb R^+$ and $\forall \delta \in \mathbb R^+$: $d(x,y)\leq\delta \implies d(f(y),f(x))\ge \epsilon_o$. For the same $\epsilon_o$, any $\delta \in \mathbb R^+$ there is $N(\delta):\forall n,m\geq N(\delta) \implies d(x_n,x_m)<\delta$ and $d(f(x_n),f(x_m))\ge \epsilon_o$. So, $f(x_n)$ is not a cauchy sequence. Which contradicts our assumption. Hence, (C) is true. Am I correct?

Answer 1

(A) Correct. Your argument is fine.

(B) Correct. Cauchy sequences with a convergent subsequence are convergent, even if the space is not complete.

(C) Correct. Let $x=\lim x_n$, $y_{2n-1}=x_n$ and $y_{2n}=x$. By (A), ${y_n}$ is Cauchy. Then ${f(y_n)}$ is Cauchy by hypothesis. Since ${f(y_{2n})}={f(x)}$, it follows by (B) that $f(y_n)$ converges and $\lim f(y_n)=f(x)$. Therefore, $f$ is continuous at $x$.

Answer 2

For (C), suppose $x_n \to x$. Then $x_n$ is Cauchy and hence so is $f(x_n)$.

We want to show that $f(x_n) \to f(x)$.

Note that the sequence $x_1,x,x_2,x,x_3,x,...$ is Cauchy from (A). Hence $f(x_1),f(x),...$ is also Cauchy and from (B) we see that $f(x_n) \to f(x)$, hence $f$ is continuous at $x$.