Question: I have a question on the example you gave about the open sets the example that you gave in the beginning of class when you started the lecture on Thursday. I'm getting confused with the part that says that A is a subset of X, then A=Y intersection of U for some open set U in X. Question what is Y and what is X. I know that Y are the intervals but, what is X? Is [0,1) open in X? NO, why not? This is where I got all confused.

Answer: OK, let's go one step at a time, using only the definitions, not using any intuition. We must become robots.

Let X = R = the set of real numbers, in the order topology

Let A=[0,1).

Your first question is, is A open in X?

The answer is no. Here's why.

Suppose temporarily that A was open in X.

Now let's remember what it means to be open in X. Well, X has the order topology. Let's remember the definition of the order topology.

First, let B be the union of three sets:

the set S1 of all open intervals (a,b), where a and b are in X and a is less than b, and

the set S2 of all half-open intervals [a,b), where a and b are in X and a is less than b and a is a minimum of X, and

the set S3 of all half-open intervals (a,b], where a and b are in X and a is less than b and b is a maximum of X.

Note that R does not have a minimum. Therefore S2 is empty. Similarly, S3 is empty. Therefore, B=S1.

The order topology is the topology generated by the basis B.

In other words, if C is any subset of X, then C is open in the order topology if and only if for every c in C, there exists an element D in B such that c is in D, and D is a subset of C. (Here we're using the definition of the topology generated by a basis.)

So, since we're assuming that A is open in X, this must mean that for every c in A, there exists an element D in B such that c is in D, and D is a subset of A.

In particular, since 0 is in A, there exists an element D in B such that 0 is in D, and D is a subset of A.

Since B=S1 and D is in B, we must have that D=(a,b) for some a, b in X with a less than b.

Since 0 is in D=(a,b), then a must be less than 0.

So then a/2 is in D. And since D is a subset of A, that implies that a/2 is in A.

But a/2 is negative, and A=[0,1), which does not contain any negative numbers.

This is a contradiction.

Therefore, A is not open in X.

Question Then is [0,1) open in Y? Yes, but why?

Let Y = [0,1) union {2}.

The topology on Y is the subspace topology inherited from X.

Let A=[0,1).

To determine whether A is open in Y, we must use the definition of the topology on Y.

In this case, the topology is the subspace topology.

In the subspace topology, A is open in Y if and only if there exists a subset U of X such that U is open in X and A equals U intersect Y.

Let U=(-1,1).

Then U is open in X. (Use the definition of the topology on X---that is, the order topology---to prove that U is open in X.)

Also, A equals U intersect Y.

Therefore, by the definition of the subspace topology, A is open in Y.

Weird---A is open in Y, but A is not open in X.

Question: Then you asked if {2} is open in Y? Yes. Why? To me it would be closed because I can't find any number that moves to my right.

Answer: Aha! Your intuition is good, but intuition can sometimes be misleading. So instead, let's be robots. As robots, we only understand definitions.

Let A={2}.

I claim that A is open in Y.

To prove that this is true, we must use the definition of the topology on Y.

The topology on Y is the subspace topology inherited from X.

In the subspace topology, A is open in Y if and only if there exists a subset U of X such that U is open in X and A equals U intersect Y.

Let U=(1,3). Then U is open in X. (Use the definition of the topology on X---that is, the order topology---to prove that U is open in X.)

Also, A equals U intersect Y.

Therefore, by the definition of the subspace topology, A is open in Y.

By the way, it is also true that {2} is closed in Y. You might try proving this.

Question Let Y = [0,1) union {2}. Show {2} Closed.

Answer To show {2} is closed in Y, we need to show Y-{2} is open. If we let U=(-1,1), U is open in X, and Y intersect U = [0,1) = Y-{2} is open. Therefore {2} is closed. QED