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**tarheelborn** ... and (2) f is continuous at a if, for every sequence $\displaystyle \{ x_n \} \in M_1$ such that $\displaystyle \displaystyle\lim_{n \to \infty} \left(x_n\right)=a$, the sequence $\displaystyle \{ f \left(x_n\right) \}$ converges to $\displaystyle f\left(a\right)$.

I need to prove $\displaystyle 2 \Longrightarrow \ 1 $. We did the first part in class:

I employ the contrapositive approach. Suppose f does not have property (1) and suppose that $\displaystyle \displaystyle \lim_{n \to \infty} \left(x_n\right)=a, \ a \ \in \ M_1$. Now if (1) does not hold, then there is $\displaystyle \epsilon > 0$ so that, $\displaystyle \forall \delta > 0$, $\displaystyle f\left(B\left[a;\delta\right]\right) \notin B\left[f\left(a\right);\epsilon\left]$. When $\displaystyle \delta = \frac{1}{n}, \ f\left(B\left[a;\frac{1}{n}\right]\right) \notin B\left[f\left(a\right);\epsilon\right]$. So in $\displaystyle B\left[a;\frac{1}{n}\right]$ there is some $\displaystyle x_n \in B\left[a;\frac{1}{n}\right]$ so that $\displaystyle f \left(x_n \right) \notin B \left[f \left(a\right);\epsilon\right].$

Now I am to my problems: 1)I need to prove that, in fact, $\displaystyle \displaystyle \lim_{n \to \infty}\left(x_n\right) = a$. Let $\displaystyle \epsilon>0$. I need to find $\displaystyle N \in \mathbb{N}$ so that, for $\displaystyle n\geq N, \ \rho\left(x_n,a\right)<\epsilon$.

I am inclined to let $\displaystyle \epsilon=1/n$, but then $\displaystyle \epsilon$ might not be an integer. So I am not sure what to do about that. Once I find my $\displaystyle \epsilon$, I simply need to finish the epsilon proof of this part.