[note: also under discussion in s.o.s. math board]
That could not be any more direct.
$\displaystyle \forall n\left[ {a - \frac{1}{n}} \right]$ is not an upper bound.
So $\displaystyle
\left( {\exists x_n \in S} \right)\left[ {a - \frac{1}
{n} < x_n \leqslant a < a + \frac{1}
{n}} \right] \to \left| {x_n - a} \right| < \frac{1}
{n} \to 0$
Is it possible to prove using only the definition of liimit?
By definition of limit,
x_n ->a iff
for ALL ε>0, there EXISTS an integer N such that n≥N => |x_n - a|< ε.
To show that a sequence converges to a certain limit, we need to FIND an integer N such that n≥N => |x_n - a|< ε. How can we construct such an N that works?
Thanks!
First construct the sequence. $\displaystyle \forall n\left[ {a - \frac{1}{n}} \right]$ is not an upper bound.
So $\displaystyle \left( {\exists x_n \in S} \right)\left[ {a - \frac{1}
{n} < x_n \leqslant a < a + \frac{1}
{n}} \right] \to \left| {x_n - a} \right| < \frac{1}
{n}$
The suppose that $\displaystyle \varepsilon > 0$. Then $\displaystyle \left( {\exists J} \right)\left[ {\frac{1}{J} < \varepsilon } \right]$.
Now $\displaystyle n \geqslant J \Rightarrow \quad \frac{1}{n} \leqslant \frac{1}{J} < \varepsilon $.
That means $\displaystyle n \geqslant J \Rightarrow \quad \left| {x_n - a} \right| < \varepsilon $.
That is the definition.
OK, so I think taking J>(1/epsilon) works and this should complete the proof.
But at the beginning of the proof, they said that ε = 1/n????? So I am a little confused...why is ε changing for different values of n? This is really werid...shouldn't ε be given to be some FIXED positive number?
When they say "ε = 1/n" at the beginning of the proof, is this the "same" ε as the one that appears at the end when we're trying to prove the limit using definition? or are they totally unrelated?
thanks!