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.

  • $\begingroup$ $C$-convergent sequences are eventually constant (i.e. has a constant tail) and of course all such sequences are convergent. Converse is not true, not all convergent sequences are eventually constant $\endgroup$ – user160738 Feb 6 '16 at 21:05

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. :)

  • $\begingroup$ So does this mean C-convergence implies convergence, but convergence does not necessarily imply C-convergence? $\endgroup$ – user57891 Feb 6 '16 at 21:00
  • $\begingroup$ Precisely, yes. In some sense "most" convergent sequences are not C-convergent. Take $(1/n)$ for instance. $\endgroup$ – Dan Feb 6 '16 at 21:00

$C$-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 $n>N$, $x_n=x$.


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