Alternative version of definition of convergence

Question

Knowing the definition of convergence of a sequence in $\Bbb R$: ($x_n$) converges to x iff $\forall \epsilon \gt 0$ $\exists N$ such that for every $n\gt N$, $d(x_n,x)\lt \epsilon$.

Consider a new definition, say $C-convergence$: ($x_n$) C-converges to x iff $\exists N$ such that for every $\epsilon \gt 0$ and every $n\gt N$, $d(x_n,x)\lt \epsilon$.

Does C-convergence imply convergence? Also, does convergence imply C-convergence? Could someone provide a proof or counterexample to solve this problem? I'm confused about switching the order of the phrases in this definition. Thanks.

Answer 1

Hint: Assume $(x_n)$ is C-convergent to $x$. Fix $n > N$. Then for all $\epsilon > 0$ we know that $0 \leq d(x_n,x) < \epsilon$. So $d(x_n,x) \in \cap_{\epsilon > 0} [0,\epsilon) = \{0\}$, which is to say that $x_n = x$. Thus a sequence $(x_n)$ is C-convergent if and only if there exists $x \in \mathbb{R}$ and $N$ such that $x_n = x$ for all $n > N$.

Perhaps this stands for constant-convergence. :)

Answer 2

$C$-convergence is much stronger than convergence. The sentence

for every $\epsilon>0$ and every $n>N$, $d(x_n,x)<\epsilon$

is the same as

for every $n>N$, $x_n=x$.