If (c) is a constant sequence in F (F is an ordered Field), then it has a limit c. HINT: for each e > 0, let K = 0.

How do I go about proving this? Any advice?

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- Apr 20th 2009, 09:40 PMjzelltLimit of Constant Sequence
If (c) is a constant sequence in F (F is an ordered Field), then it has a limit c. HINT: for each e > 0, let K = 0.

How do I go about proving this? Any advice? - Apr 20th 2009, 11:34 PMGammaDefinition
A sequence $\displaystyle \{x_n\}_{n=1}^{\infty}$ is said to converge to c iff for any $\displaystyle \epsilon > 0$, $\displaystyle \exists K\in \mathbb{N}$ so that $\displaystyle \forall n > K, |x_n - x_K| < \epsilon$. In this case we say the $\displaystyle x_n \rightarrow c$

In your case your function is constant and is c for every n, so $\displaystyle |x_n-x_0|=|c-c|=|0|=0 < \epsilon$ for all $\displaystyle \epsilon >0$ this proves your sequence converges to c by choosing K=0 in the definition.