5

To prove this, I want to show that an arbitrary intersection of closed sets is closed and an arbitrary intersection of bounded sets is bounded. I know how to prove the first part, but I'm not sure how to rigorously show that an intersection of bounded set is bounded.

This question comes from a real analysis course, not a general topology course.

Moz
  • 385
  • 4
    Bounded means contained within a sphere. If $A$ is a contained within a sphere, so is any subset of $A$. – Paul Sinclair Sep 23 '15 at 03:32
  • Ok, I wasn't sure if it was anything more than that -- i.e. I felt I was missing rigor, but thanks. – Moz Sep 23 '15 at 03:33
  • Closed and bounded sets need not be compact. – John Douma Sep 23 '15 at 03:35
  • 2
    Rigor does not mean "complicated". What I said is completely rigorous. – Paul Sinclair Sep 23 '15 at 03:35
  • Closed and bounded sets need not be compact? The definition of compactness I use is that a compact set is bounded and closed. – Moz Sep 23 '15 at 03:37
  • @JohnDouma - it is apparent to me that Moz is studying real or complex analysis, and has not yet had exposure to general topology. So in his case, closed and bounded is compact. – Paul Sinclair Sep 23 '15 at 03:38
  • 2
    @Moz - in general, compact means that for every covering of the set by a collection of open sets, there is a finite subcollection of those open sets that also covers the compact set. In the Reals or complex numbers this is equivalent to closed and bounded. But in general, it is more stringent (and can even be applied where "bounded" has no meaning). – Paul Sinclair Sep 23 '15 at 03:39
  • Thanks, I will make a note in my question that this is from an analysis class, not a topology class. – Moz Sep 23 '15 at 03:42
  • @Moz Okay, then I also give a +1 to Paul's hint. A subset of a bounded set must be bounded because it is smaller and $A\cap X\subset A$ for all $X$. – John Douma Sep 23 '15 at 03:42
  • 1
    As an aside, this result is actually very easy to prove from the general definition, since the intersection of closed sets is always closed (a statement which itself can be considered part of the definition of a topology). The intersection is a closed subset of a compact set. Any cover of the closed subset can be extended with the complement of the closed set to cover the entire compact set. A finite subcover of that set with the complement removed has to be a finite subcover of the original cover of the closed set. QED – Paul Sinclair Sep 23 '15 at 03:49

2 Answers2

5

The proof is somewhat dependent on the definitions you use.

The simple (as you basically have only one possible definition) part is the intersection of bounded sets. You either define bounded as having a radius such that every member of the set has absolute value (or norm) less or equal to the radius, or more general that the distance of any two points in the set have distance $d(x,y)$ less or equal to the radius (I use the latter), I don't require the radius to be the smallest bound here (if one bound exists there exists an infimum of them which is also a bound by the axiom of largest lower bound):

Let $B_j$ be bounded sets which means that there exists a radius $r_j$ such that $d(x,y)\le r_j$ for all $x,y\in B_j$. Now consider the intersection $B = \bigcap B_j$, then we have that $d(x,y)\le r_o$ for any $x,y\in B_o$, but since $B \subseteq B_o$ we have for any $x, y\in B$ that $x,y\in B_o$ and therefore $d(x,y)\le r_o$, so $r_o$ is a radiuos of $B$.

To prove that intersection of closed sets is closed we have to do that differently depending on the definition:

  • Closed being that the complement is open: Use deMorgan theorem and use that $\overline{F_j}$ is open, therefore $\bigcap F_j = \overline{\bigcup \overline {F_j}}$, but the union of open sets is open so the intersection is the complement of an open set.
  • Closed being containing all it's limit points. Suppose we have $a$ being a limit point of $F =\bigcap F_j$, then there exists a sequence $a_k\in F$ such that $\lim_{k\to\infty}a_k = a$, but since $F_j\subseteq F$ we have that $a_k$ is a sequence in $F_j$ and therefore $a\in F_j$ as $F_j$ is closed. And since $a\in F_j\subseteq F$ we have $a\in F$. So $F$ contains all its limit ponts.

Also for compactness it's a matter of definition:

  • Compact being closed and bounded: The intersection of closed is closed, and intersection of bounded is bounded. Therefore intersection of compact is compact.
  • Compact being that open cover has a finite subcover: This is a lot trickier (and may be out of your scope), I will need to use more assumptions here. Suppose we have a open cover of $F \subset \cup \Omega_j$. We extend this to a cover of any $F_o$ by observing that for each $\phi\in F_o\setminus F$ we have an open set $\Omega_\Phi$ that doesn't intersect $F$ (assuming the space is regular Hausdorff - for example the set of real numbers). Now we have an open cover of $F_o$ which have a finite subcover, and excluding the ammended sets to cover the whole $F_o$ will make it a cover of $F$.
skyking
  • 16,654
-2

The finite subcover produced by a subcover of one of the compact sets also covers the intersection.

gabrielchua
  • 163
  • This is wrong. Covering an intersection is a lot easier than covering every single of its constituents. – J. De Ro Mar 21 '20 at 10:05