Thread: Continuous and Unbounded Function

If a set S contains an unbounded sequence , show that the function defined by for all in , is continuous, but unbounded. and If a set S contains a sequence that converges to a point in S, show that the function defined by for all in , is continuous and unbounded

All I could figure out was that
for the first case, contains an unbounded sequence. We can set, for every index n, and thus every subsequence of this sequence is unbounded and fails to converge.

How can it be proven for the defined function?

Also in the second part, has a sequence that converges to 0 that is not in S.

Again I am not being able to do this for the given function.

Any suggestions??

For the first part, we know there is some unbounded sequence , so the sequence is obviously unbounded, meaning f is unbounded.

To show that it's continuous, use the sequence definition of continuity; a function is continuous if implies . That should be obvious since f is the identity function.

The second part is similar.