# Thread: show f is bounded two ways..?

1. ## show f is bounded two ways..?

Hey guys, so this question was posted a while ago by someone so I don't know if this is reposting, but my question about it is different.

Suppose f: K->(- $\infty$, $\infty$), K is compact, and f has a finite limit at each point of K, but may not be continuous on K. Show that f is bounded in two ways: (i) by using the definititon of compactness in terms of open covers, and (ii) by using the sequential characterization of compactness. Is the same conclusion valif if we drop the assumption that the limit of f is finite?

For some reason I can't find these specific definitions in the book I'm using, but I've come acrossed problems that deal with them, similar to this one. I kinda want to battle out the problems I've been working on on my own but I feel like seeing the proof for this problem for part one and part two would give me a better understanding of what's going on here and and lead me to figure out my stuff. Thanks!

2. Well, what definition of "compact" does your book have?

The first says that a set, A, is compact if and only if every open cover of A has a finite subcover. That is, given any collection of open sets such that every point of A is in at least one of the open sets, there exist a finite sub-collection of those same open sets such that every point of A is in at least one of the sets in the subcollection.

The second says that a set, A, is compact if and only if every infinite sequence $\{a_n\}$ of points in A converges to a point in A.

To use the first, for example, since f has a finite limit at every point, x, given any $\epsilon> 0$, there exist a number, $\delta_x> 0$ for each x such that if $d(y,x)< \delta_x$ then $|f(x)- f(y)|< \epsilon_x$. let $N_\delta_x(x)$ be the neighborhood about point x with radius $\delta_x$. These form an open cover of K. Since K is compact, there exist a finite sub-cover. Now we have a finite number of points, $x$ such that every point in K is in some $N_x(x)$ which means that for y in $N_x(x)$, $f(x)-\epsilon$< f(y)< f(x)+ \epsilon[/tex]. Since the set of "x"s is finite, there is a largest such " $f(x)+ \epsilon$" and a smallest such " $f(x)- \epsilon$" which are bounds for f(K).

3. Originally Posted by HallsofIvy
Well, what definition of "compact" does your book have?

The first says that a set, A, is compact if and only if every open cover of A has a finite subcover. That is, given any collection of open sets such that every point of A is in at least one of the open sets, there exist a finite sub-collection of those same open sets such that every point of A is in at least one of the sets in the subcollection.

The second says that a set, A, is compact if and only if every infinite sequence $\{a_n\}$ of points in A converges to a point in A.

To use the first, for example, since f has a finite limit at every point, x, given any $\epsilon> 0$, there exist a number, $\delta_x> 0$ for each x such that if $d(y,x)< \delta_x$ then $|f(x)- f(y)|< \epsilon_x$. let $N_\delta_x(x)$ be the neighborhood about point x with radius $\delta_x$. These form an open cover of K. Since K is compact, there exist a finite sub-cover. Now we have a finite number of points, $x$ such that every point in K is in some $N_x(x)$ which means that for y in $N_x(x)$, $f(x)-\epsilon$< f(y)< f(x)+ \epsilon[/tex]. Since the set of "x"s is finite, there is a largest such " $f(x)+ \epsilon$" and a smallest such " $f(x)- \epsilon$" which are bounds for f(K).
Oh ok. My book's def of compact is basicially the same now that I found it. It for some reason is stated very complex that I didn't even recognize it. But now with the def you gave I can figure out what mine is saying now although it passes subtly over open covers which is why I was having trouble finding it and was getting confused. However, now with the second way to show it am I basically showing that the sequence converges and if it does then is bounded? Would I show that by showing the sequence is cauchy? And thanks again u really cleared up that def for me!