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Here is a problem I am working on that has begun to really frustrate me but maybe I am over thinking it:

"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."

then they continue...

"If a set $S$ contains a sequence that converges to a point $x_0$ not in $S$, show that the function $f: S \rightarrow R$, defined by $f(x) = 1/|x - x_0|$ for all $x$ in $S$, is continuous but not bounded."

I split it up because it makes more sense to tackle the first part and THEN do the second part. So far I have been thinking as follows:

So $S$ contains some unbounded sequence. Okay. If $S$ were the naturals, then clearly the unbounded sequence $\{1,\ 2,\ 3,\ 4,\ \cdots\}$ is contained in the naturals (for example). Does the fact that $S$ contains some unbounded sequence imply that $S$ is infinite? Moving along, I was starting to think that okay I need to show that the function is continuous. How do I do this? Well to show that something is continuous at a point $x$, you take any sequence that converges to that point and if the image sequence $\{f(x_n)\}$ converges to $f(x)$.

Because the sequence contained in $S$ is unbounded, I had the idea to make a sequence $\{y_n\}$ that converges to infinity (haha) and then to show that the function is continuous at infinity. My image sequence $\{f(y_n)\}$ then converges to $f(\infty)$ which is just infinity. Since my function is $f(x) = x$, have I just shown that the function is continuous at infinity?

Anyone have any recommendations for Real Analysis books, preferably with solutions available? Royden's is quite good, as is... Ramachandran or whatever. Advanced Calculus by Fitzpatrick is so horrible, the proofs make huge logical jumps and despite a very high A in my intro to proofs class with an amazing professor, this class is a nightmare.

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An unbounded subset of $\Bbb R$ clearly has to be infinite: every finite set is bounded below by its smallest element and above by its largest element. However, you don’t need this to conclude that $f$ is unbounded if $S$ is: $f[S]=S$ by the definition of $f$. The harder part is showing that $f$ is continuous, but it’s very little harder. Suppose that $\langle x_n:n\in\Bbb N\rangle$ is a sequence in $S$ converging to some $x\in S$. Does $\langle f(x_n):n\in\Bbb N\rangle$ converge to $f(x)$? Sure, because $\langle f(x_n):n\in\Bbb N\rangle=\langle x_n:n\in\Bbb N\rangle$, and $f(x)=x$. Thus, $f$ takes convergent sequences and their limits in $S$ to convergent sequences and their limits in $\Bbb R$ and so is continuous.

If a set $S$ contains a sequence that converges to a point $x_0$ not in $S$, show that the function $f:S\to\Bbb R$, defined by $$f(x) = \frac1{|x - x_0|}$$ for all $x\in S$, is continuous but not bounded.

You didn’t ask about this part of the question, but here are a couple of hints. To show that $f$ is unbounded, consider what it does to the points of a sequence converging to $x_0$. Continuity of $f$ can be shown very straightforwardly by considering what $f$ does to convergent sequences in $S$ with limits in $S$.

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