Continuous and Unbounded Function

**harish21** · Mar 24th 2010, 11:08 PM

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

All I could figure out was that
for the first case, $S \equiv [0, \infty)$ contains an unbounded sequence. We can set, for every index n, ${a_n} = n$ 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, $S \equiv (0, 2]$ has a sequence $\frac{1}{n}$ that converges to 0 that is not in S.

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

Any suggestions??

**roberto** · Mar 25th 2010, 12:06 AM

For the first part, we know there is some unbounded sequence $\{s_n\} \subset S$ , so the sequence $f(s_n) = s_n$ is obviously unbounded, meaning f is unbounded.

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

The second part is similar.

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