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11. Studying and Problem-Solving

≪ 10. Connectedness | Table of Contents | 12. Derivatives ≫

We had a midterm last week, and something that has come to our attention is that some of you may be using study habits from different classes that don’t work as well in mathematics. We have recommended several textbooks as references for this course, but we do not recommend just reading a textbook as your main method of studying. Rather, it’s a better use of your time to be attempting exercises: Professor Greene provides plenty of practise exercises, and I try to provide some good exercises to go with our discussions every week. Additionally, your textbooks have a lot of great exercises in them.

A lot of math is just building the crystalline knowledge in theorems and definitions, but an even bigger part of it is understanding how to use them! This is why doing plenty of exercises is so important, and this isn’t a skill that can be built by reading alone.

Of course, some exercises are very difficult, and it’s easy to get stuck. If this happens, you should refer to the textbook — sometimes, seeing a proof of a similar theorem or seeing certain propositions again provides the inspiration needed to get unstuck. If you find yourself spending way too much time on an exercise, don’t feel ashamed to just look it up or ask someone for help! You should not try to do everything yourself. That is too much of a burden for an individual mortal to shoulder.

Due to the holiday and the slight lull in new content, we’ll be reviewing some of the older content (continuity, compactness, connectedness, etc.) with some practise exercises! Some of these will be taken from or inspired by past basic exams. We’ll go through the first few together and identify some helpful strategies along the way.

Let’s warm up with a problem about relative topologies.

Exercise 1.

Let \(\left( X,d \right)\) be a metric space and let \(S\) be any subset of \(X\). Show that \(U\subseteq S\) is open in \(S\) if and only if there exists some \(U’\subseteq X\) open in \(X\) such that \(U=U’\cap S\). Likewise, show that \(C\subseteq S\) is closed in \(S\) if and only if there exists some \(C’\subseteq X\) closed in \(X\) such that \(C=C’\cap S\).

Exercise 2. (Fall 2014, Problem 2)

Let \(A,B\) be two closed subsets of \(\mathbb{R}^n\) such that \(A\cup B\) and \(A\cap B\) are both connected. Prove that \(A\) is connected.

Let’s walk through a solution of this one together. Let’s start by remembering the two different ways to think about connectedness:

We should also think about the assumptions in the problem statement. At some point in our solution, we will have to use the fact that \(A\) and \(B\) are closed, as well as the fact that their union and intersection are both connected. If any of these three assumptions fail, then \(A\) need not be connected. A sub-exercise is to think of some examples!

We can now attempt a solution. I can’t immediately see any reason to prefer one version of connectedness over the other, so I’ll default to the first definition we gave. Let \(C\subseteq A\) be a nonempty clopen subset; we wish to show that \(C=A\).

Our goal is to now relate the clopen-ness of \(C\) to the connectedness of \(A\cup B\) and \(A\cap B\). After all, connectedness by definition tells us about what clopen subsets should look like. But in order to do that, we need to use the fact that \(C\) is clopen in A to demonstrate that \(C\cap \left( A\cap B \right)\) and \(C\cap \left( A\cup B \right)\) are clopen in their respective subspaces.

We know \(A\cap B\) is connected, and we know that \(C\cap \left( A\cap B \right)\) is clopen in \(A\cap B\). Thus we know that \(C\cap \left( A\cap B \right)\) is either \(\varnothing\) or \(A\cap B\). This is a dead end — we can’t rule out either case a priori. However, what we do know is that the intersection should be \(A\cap B\). After all, we’re supposed to conclude that \(C=A\).

We do know that \(A\cup B\) is connected, but does \(C\) remain clopen in \(A\cup B\)? On one hand, \(C\) does remain closed. This is because \(A\) is closed in \(\mathbb{R}^n\) by assumption, and since \(C\) is closed in \(A\), \(C=C’\cap A\) for some \(C’\subseteq \mathbb{R}^n\) closed (but \(C’\) may not be contained in \(A\)). But the intersection of closed sets is closed, so \(C=C’\cap A\) is closed in \(\mathbb{R}^n\). It then follows that \(C\cap \left( A\cup B \right)\) is closed in \(A\cup B\).

Unfortunately \(C\) may not remain open in \(A\cup B\). In fact, \(A\) is clopen in \(A\) yet should not be clopen in \(A\cup B\) unless \(A=A\cup B\)! So this tells us that something has gone wrong. The problem is that the “boundary” of \(A\) could land in the “interior” of \(B\). While this boundary does not pose any problems for open-ness in \(A\), it does pose a problem for open-ness in \(B\). Draw a picture to see why!

Morally speaking, the only way for \(C\) to remain open in \(A\cup B\) is if \(C\) avoids the “insides” of \(B\) altogether. Ah! Notice that we mentioned earlier that \(C\cap \left( A\cap B \right)=\varnothing\) should not be possible.

Putting things together, here’s our chain of reasoning:

The point is that if \(C\) is a nonempty clopen subset of \(A\), it has to contain \(A\cap B\). But then \(A\setminus C\) is also a clopen subset of \(A\), which must either be empty or also contain \(A\cap B\). This forces \(A\setminus C\) to be empty, and we conclude that \(C\) must be all of \(A\).

The only missing ingredient is proving that if \(C\subseteq A\) is open and \(C\cap \left( A\cap B \right)=\varnothing\), then \(C\) is open in \(A\cup B\). I’ll let you handle that part!

Hopefully this problem-solving process makes sense. I wrote this up while solving the problem real-time, so this hopefully accurately reflects some of the ways in which I think about and approach exercises like this. We could have just as well used the disjoint open cover definition of connectedness, and it only slightly changes the ideas in the proof. It would be good to go back and work things out!

At the end of the day, we didn’t use any obscure theorems or extremely difficult lines of reasoning. It was mostly tossing around definitions and seeing what we could and could not say, sometimes using our intuition to guide us towards what “should” be said!

Here’s a bunch of other exercises for your practise. They follow the same theme of, if you can push the definitions around, you will eventually solve the problem. Knowing exactly how to push things around and what can be said is just something that comes with practise.

Exercise 3.

Let \(\left( X,d_X \right)\) and \(\left( Y, d_Y \right)\) be metric spaces. Let \(f:X\to Y\) be a function. Show that \(f\) is continuous if and only if \(f\) is continuous on every compact subset of \(X\) (i.e., the restriction \(f:K\to Y\) is continuous whenever \(K\subseteq X\) is compact).

Hint
Pick some \(x_0\in X\) where \(f\) fails to be continuous. Find a sequence \(\left\lbrace x_n \right\rbrace\) in \(X\) that “realises” this failure of continuity. If \(x_n\to x_0\) (which it should), you can show that \(\left\lbrace x_0,x_1,x_2,x_3,\ldots \right\rbrace\) (i.e., the sequence plus its limit point) is compact.

Exercise 4.

Last week, we gave an example of a connected space that was not path-connected. Under certain circumstances, however, we can say that the idea of topological connectedness and path-connectedness coincide:

Let \(U\subseteq \mathbb{R}^n\) open and connected. Show that \(U\) is path-connected.

Hint
Pick any \(x_0\in U\), and let \(C\) be the set of points in \(U\) that can be connected to \(x_0\) with a continuous path. Show that \(C\) is both open and closed.

Exercise 5. (Spring 2016, Problem 6)

A metric \(d\) in a metric space \(\left( X,d \right)\) is said to be an ultrametric if for every \(x,y,z\in X\), \[d\left( x,y \right)\leq \max \left( d\left( x,z \right), d\left( z,y \right) \right).\] Prove that, in this metric, every open ball is closed and every closed ball is open.

Exercise 6. (Spring 2017, Problem 12)

Let \(K\subseteq \mathbb{R}^n\) be compact. Suppose that for every \(\epsilon>0\) and every pair \(a,b\in K\), there is an integer \(n\geq 1\) and a sequence of points \(x_0,\ldots, x_n\in K\) such that \(x_0=a\), \(x_n=b\), and \(\left\lVert x_k-x _{k-1} \right\rVert<\epsilon\) for every \(1\leq k\leq n\).

  1. Show that \(K\) is connected.
  2. Show by example that \(K\) may not be path-connected.