# Thread: there exist a sequence x_n E S s.t. x_n -> sup S

1. ## there exist a sequence x_n E S s.t. x_n -> sup S

[note: also under discussion in s.o.s. math board]

2. It is simple: $\left( {\frac{1}{n}} \right) \to 0$ implies that $\left| {a - x_n } \right| \to 0$.
It is the ole squeeze play.

3. Yes, but how can we prove directly from the definition that my constructed sequence x_n converges to a? (how can we choose an N that will work in the definition of limit?)

4. Originally Posted by kingwinner
Yes, but how can we prove directly from the definition that my constructed sequence x_n converges to a? (how can we choose an N that will work in the definition of limit?)
That could not be any more direct.

$\forall n\left[ {a - \frac{1}{n}} \right]$ is not an upper bound.
So $
\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$

5. 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!

6. Originally Posted by kingwinner
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?
First construct the sequence. $\forall n\left[ {a - \frac{1}{n}} \right]$ is not an upper bound.
So $\left( {\exists x_n \in S} \right)\left[ {a - \frac{1}
{n} < x_n \leqslant a < a + \frac{1}
{n}} \right] \to \left| {x_n - a} \right| < \frac{1}
{n}$

The suppose that $\varepsilon > 0$. Then $\left( {\exists J} \right)\left[ {\frac{1}{J} < \varepsilon } \right]$.
Now $n \geqslant J \Rightarrow \quad \frac{1}{n} \leqslant \frac{1}{J} < \varepsilon$.

That means $n \geqslant J \Rightarrow \quad \left| {x_n - a} \right| < \varepsilon$.
That is the definition.

7. Originally Posted by Plato
First construct the sequence. $\forall n\left[ {a - \frac{1}{n}} \right]$ is not an upper bound.
So $\left( {\exists x_n \in S} \right)\left[ {a - \frac{1}
{n} < x_n \leqslant a < a + \frac{1}
{n}} \right] \to \left| {x_n - a} \right| < \frac{1}
{n}$

The suppose that $\varepsilon > 0$. Then $\left( {\exists J} \right)\left[ {\frac{1}{J} < \varepsilon } \right]$.
Now $n \geqslant J \Rightarrow \quad \frac{1}{n} \leqslant \frac{1}{J} < \varepsilon$.

That means $n \geqslant J \Rightarrow \quad \left| {x_n - a} \right| < \varepsilon$.
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!

8. Originally Posted by kingwinner
OK, so I think taking N>(1/epsilon) works.

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?
There are many students who are so profoundly confused by $\varepsilon$-limit proofs, they simply require intensive one-on-one instruction/tutorial. This may be the case for you.