# Math Help - Sequence

1. ## Sequence

Suppose $f$ is a real valued function whose domain contains the interval $[a,b]$. Assume that $f$ is unbounded on $[a,b]$.

(a) Prove that there exists a sequence $(y_n)$ in $[a,b]$ that converges to $y \in [a,b]$ such that for every $n \in \mathbb{N}$, $|f(y_n)| > n$.

Isn't this just the definition of an unbounded sequence? Do we have to actually construct one? What would be the easiest way to do this?

(b) Using this unbounded sequence how would we show that the set of Riemann sums of $f$ corresponding ot $P$ is an unbounded set of real numbers?

2. GENERALLY speaking ,usually in forming sequences in analysis there are two basic ways:

1)By using the axiom of Choice.

2)By using the inductive method

The following steps are a summary of the process in forming the sequence in concern,by using the axiom of Choice.

step 1) PROVE: For all nεΝ there exists an ,xε [a,b] and such that |f(x)|>n,by using the fact that f is unbounded over [a,b]

step 2) Form the sets $A_{n}$={x:xε[a,b],|f(x)|>n,nεN}

Step 3) By using step 1 prove : For all neΝ, $A_{n}\neq\emptyset$.

step 4) Now use the axiom of Choice to form a sequence { $x_{n}$} in [a,b] and such that |f( $x_{n}$)|>n ,nεN.

Step 5) By the Bolzano Weiestrass theorem ,since the sequence { $x_{n}$} is bounded in [a,b] ,there exists an accumulation point ,v in [a,b].

step 6)Prove:For all nεN ,THERE exists a , $y\neq v$,yε{ $x_{n}:n\in N$} and such that |y-v|<1/n,by using the fact that v is an accumulation point.

step 7)Form the sets, $S_{n}$ ={y:|y-v|<1/n,nεΝ, $y\neq v$,yε{ $x_{n}$:nεN}}

Step 8) Prove: for all nεN $S_{n}\neq\emptyset$ by using step 6.

step 9) Use again the axiom of Choice to form the sequence { $y_{n}$} in [a,b],such that |f( $y_{n}$)|>n , $y_{n}\neq v$ and | $y_{n}$-v|<1/n for all nεΝ

step 10) prove , $\lim_{n\rightarrow\infty}{y_{n}}=v$.

ALTERNATIVELY,by using the inductive process we can say:

After we have proved that ,for all n belonging to natural Nos ,there exists an ,x belonging to [a,b] and such that |f(x)|>n.

For n= 1 ,we pick $x_{1}$,such that |f( $x_{1}$)|>1.

For n=2 ,we pick $x_{2}>x_{1}$,such that |f( $x_{2}$)|>2.

Suppose we have found:

$x_{1} all in [a,b] and such that |f( $x_{n}$)|>n .

We select $x_{n+1}>x_{n}$.We can thus proceed inductively to construct the sequence { $x_{n}$},which is increasing and bounded from above by ,a.Hence there exists a,y in [a,b] and such, $\lim_{n\rightarrow\infty}{x_{n}}=y$.

So we have proved the existence of a sequence { $x_{n}$} in [a,b] such that |f( $x_{n}$|>n and $\lim_{n\rightarrow\infty}{x_{n}}=y$