[Topology] Find the open sets in the subspace topology

[Arian.D]

Homework Statement

Suppose that [itex](X,\tau)[/itex] is the co-finite topological space on X.
I : Suppose A is a finite subset of X, show that [itex](A,\tau)[/itex] is discrete topological space on A.
II : Suppose A is an infinite subset of X, show that [itex](A,\tau)[/itex] inherits co-finite topology from [itex](X,\tau)[/itex].

The Attempt at a Solution

I: Well, first I show that [itex]\forall a\in A: (A^c \cup \{a\})\in \tau[/itex]. I need to show that [itex] (A^c \cup \{a\})^c [/itex] is finite because I want it to be in [itex]\tau[/itex]. But:
[itex] (A^c \cup \{a\})^c = A \cap (X-\{a\}) = A - \{a\}[/itex], but since A is finite, if we remove one of its element, the remaining set is again finite. Therefore [itex]\forall a\in A: (A^c \cup \{a\})\in \tau[/itex]
Now since all the open sets in the subspace topology are the intersection of [itex]A[/itex] with something in [itex]\tau[/itex], we see that [itex]\{a\} = A \cap (A^c \cup \{a\})[/itex] is open in [itex](A,\tau)[/itex]. That means the topology induced on A will be the discrete topology.

II: What should I do for the second part? Any helps would be appreciated.

 

[Arian.D]

Another problem:

Suppose that Z is a subspace of Y, and Y is a subspace of X, show that Z is a subspace of X.

Well, please confirm if my argument is correct.

Since Z is a subspace of Y, therefore any open set in Z must be of the form [itex] Z \cap U[/itex] where U belongs to the topology defined on Y, but since Y is a subspace of X, then U, again, is of the form [itex] Y \cap O [/itex] where O belongs to the topology on X, therefore for any open set A in Z, we'll have [itex] Z = Z \cap U = Z \cap ( Y \cap O) = Z \cap O [/itex] where O is an open set in the topological space X.

 

[micromass]

I: Well, first I show that [itex]\forall a\in A: (A^c \cup \{a\})\in \tau[/itex]. I need to show that [itex] (A^c \cup \{a\})^c [/itex] is finite because I want it to be in [itex]\tau[/itex]. But:
[itex] (A^c \cup \{a\})^c = A \cap (X-\{a\}) = A - \{a\}[/itex], but since A is finite, if we remove one of its element, the remaining set is again finite. Therefore [itex]\forall a\in A: (A^c \cup \{a\})\in \tau[/itex]
Now since all the open sets in the subspace topology are the intersection of [itex]A[/itex] with something in [itex]\tau[/itex], we see that [itex]\{a\} = A \cap (A^c \cup \{a\})[/itex] is open in [itex](A,\tau)[/itex]. That means the topology induced on A will be the discrete topology.

Looks ok.

II: What should I do for the second part? Any helps would be appreciated.

So take an open set in A. Show that it's complement is finite. What do you know about open sets in A? Relate it to open sets in X.

Another problem:

Suppose that Z is a subspace of Y, and Y is a subspace of X, show that Z is a subspace of X.

Well, please confirm if my argument is correct.

Since Z is a subspace of Y, therefore any open set in Z must be of the form [itex] Z \cap U[/itex] where U belongs to the topology defined on Y, but since Y is a subspace of X, then U, again, is of the form [itex] Y \cap O [/itex] where O belongs to the topology on X, therefore for any open set A in Z, we'll have [itex] Z = Z \cap U = Z \cap ( Y \cap O) = Z \cap O [/itex] where O is an open set in the topological space X.

Good.

 

[Arian.D]

So take an open set in A. Show that it's complement is finite. What do you know about open sets in A? Relate it to open sets in X.

I only know that open sets in A are the intersection of A with an open set in the original topology. Well, if B is a proper subset of A which is co-finite, then A - B is finite, does that mean X-B is also finite? Not necessarily. but I must show that for some subsets of A that are co-finite, there exists an open set in the original topology on X that the intersection of that open set and A is equal to the subset of A that we've chosen.

A little practical help would be very much appreciated. I know what I need to do, but a small hint is needed I guess.

 

[micromass]

You should prove two things:

1) If B is a finite subset of A, then B is closed. Can you write B as [itex]B=A\cap F[/itex] with F closed in X?

2) If B is closed, then B is finite. But you can write B as [itex]B=A\cap F[/itex] with F closed. Can you deduce that B is finite?

 

[Arian.D]

You should prove two things:

1) If B is a finite subset of A, then B is closed. Can you write B as [itex]B=A\cap F[/itex] with F closed in X?

well, let me see if I've understood you correctly. I know what you're trying to get at though, you're using a theorem (or the definition of subspace topology using closed sets). For example I could define [itex] F = B \cup C[/itex] where C is a finite subset of [itex]A^c[/itex], then [itex]B = A \cap F[/itex] where F is closed in X, then B is closed in the subspace topology. right?

2) If B is closed, then B is finite. But you can write B as [itex]B=A\cap F[/itex] with F closed. Can you deduce that B is finite?

Sure, since F is closed, F is finite, and so is B. Right?

 

[micromass]

well, let me see if I've understood you correctly. I know what you're trying to get at though, you're using a theorem (or the definition of subspace topology using closed sets). For example I could define [itex] F = B \cup C[/itex] where C is a finite subset of [itex]A^c[/itex], then [itex]B = A \cap F[/itex] where F is closed in X, then B is closed in the subspace topology. right?

Why not take F=B??

 

[Arian.D]

Why not take F=B??

That must work as well. No? Unless I'm missing something.
I got pretty confused with open sets, it seemingly appears to be easier to work with closed sets.

 

[micromass]

All of it looks correct.
Sometimes it's easier to work with closed sets instead of open sets...

 

[Arian.D]

Nice. Thanks micromass.
Could I reason exactly the same way if I replace co-finite topology with co-countable topology on X? Since nothing would change, the method of reasoning will remain the same. Right?

 

[micromass]

Right! It would be very similar.