Thread: Continuous and Unbounded Function

If a set S contains an unbounded sequence , show that the function $\displaystyle f:S \rightarrow R$ defined by $\displaystyle f(x)=x$ for all $\displaystyle x$ in $\displaystyle S$, is continuous, but unbounded. and If a set S contains a sequence that converges to a point $\displaystyle x_0$ in S, show that the function $\displaystyle f:S \rightarrow R$ defined by $\displaystyle f(x) = \frac{1}{x-x_0}$ for all $\displaystyle x$ in $\displaystyle S$, is continuous and unbounded

All I could figure out was that
for the first case, $\displaystyle S \equiv [0, \infty)$ contains an unbounded sequence. We can set, for every index n, $\displaystyle {a_n} = n$ 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, $\displaystyle S \equiv (0, 2]$ has a sequence $\displaystyle \frac{1}{n}$ 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 $\displaystyle \{s_n\} \subset S$, so the sequence $\displaystyle f(s_n) = s_n$ is obviously unbounded, meaning f is unbounded.

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

The second part is similar.