Question

Show that Xn n in N has a sub sequence Xnk k in N with Xnk > k for all k in N

Answer 1

Ok.

Answer 2

Since \( x_n \) is unbounded, then for for every \(k >0\) there is an element \( x_{n_k} >k\). In fact you can construct an increasing sub sequence ( x_{n_k} >k\). You can chose
\( x_{n_k} >(k , x_{n_{k-1}}),\)

Answer 3

I don't understand :\

Answer 4

I don't understand what the k represents, like I can't wrap my head around it

Answer 5

this is not the proof just explanation

the "main" series:
\[x_{n} , n \in N\]
now the k thing:
\[n_{k} , k \in N\]
this is a series of NATURAL NUMBERS (the nk's)

so
\[x_{n_{k}}\]
is the sub sequence

Answer 6

sorry i wrote series and meant sequence