"Show B is closed if and only if...." (More Topology....)

[sa1988]

Homework Statement

Let $A$ be a subspace of a topological space $X$, and let $B\subset A$.

Show that $B$ is closed if and only if there exists a closed subset $C \subset X$ such that $B = C \cap A$

Homework Equations

The Attempt at a Solution

So I've started by just drawing the two cases in which $B$ is said to exist. It doesn't really help with much but it's a way for me to see clearly what I'm supposed to be working with.

I believe the right hand image demonstrates the 'only if' part of the statement, where we have $B \subset A \subset X$ which is open by definition. (Since the complement of $A$ is $X$ which is open because it's a topological space, thus $A$ is closed, and then the same applies to show that $B$ is open since the complement of $B$ is closed).

So $B$ can only possibly be closed in the other case, shown in the left hand image, which is the 'if' part. From this we can see that $\exists$ neighbourhoods $N$ of $n \in B$ where $N \subset B$ and $N \subset C$, so $N \subset B \cap C$

But if $N \subset B \cap C$, this means $B$ contains elements with neighbourhoods that go beyond its boundaries into the closed set $C$.

Hence $B$ must be closed too, in the case where $B = A \cap C$, and only this case.

Is this proof heading in the right direction? I fear I may have been a bit too hand-wavy with the 'only if' part.

Thanks.

 

[fresh_42]

Homework Statement

Let $A$ be a subspace of a topological space $X$, and let $B\subset A$.

Show that $B$ is closed if and only if there exists a closed subset $C \subset X$ such that $B = C \cap A$

Homework Equations

The Attempt at a Solution

So I've started by just drawing the two cases in which $B$ is said to exist. It doesn't really help with much but it's a way for me to see clearly what I'm supposed to be working with.

I believe the right hand image demonstrates the 'only if' part of the statement, where we have $B \subset A \subset X$ which is open by definition. (Since the complement of $A$ is $X$ which is open because it's a topological space, thus $A$ is closed, and then the same applies to show that $B$ is open since the complement of $B$ is closed).

So $B$ can only possibly be closed in the other case, shown in the left hand image, which is the 'if' part. From this we can see that $\exists$ neighbourhoods $N$ of $n \in B$ where $N \subset B$ and $N \subset C$, so $N \subset B \cap C$

But if $N \subset B \cap C$, this means $B$ contains elements with neighbourhoods that go beyond its boundaries into the closed set $C$.

Hence $B$ must be closed too, in the case where $B = A \cap C$, and only this case.

Is this proof heading in the right direction? I fear I may have been a bit too hand-wavy with the 'only if' part.

Thanks.

I'm not quite sure what this problem is about. Since it is true, it is important to separate what is known from what has to be shown.

Let $A$ be a subspace of a topological space $X$, and let $B\subset A$.

So far we have only sets: $B \subseteq A \subseteq X = (X,\tau)$. One of them with a topology, and only one, namely $X$.

Show that $B$ is closed ...

Here is the first sloppy notation. It makes a difference, whether $B$ is closed in $(X,\tau)$ or closed in $A$, for which we have strictly speaking no topology yet. Thus we cannot decide, what is meant by $B$ is closed. Remember that $[0,\frac{1}{2})$ is a open set in $[0,1] \subseteq \mathbb{R}$ with the (induced) subset topology in $[0,1]$, whereas it is not open in $\mathbb{R}$. This shows that it is crucial to define where a set is regarded as open. In our case as we regard closed sets, we might change the example a little bit and define $(0,1) =: A \subseteq \mathbb{R}$ and $B=(0,\frac{1}{2}]$. Then $B \subseteq A$ is closed and $B \subseteq \mathbb{R}$ is not!

To proceed, let's make as two assumptions:

1. $B$ shall be closed in $A$.
2. But therefore we need a topology of $A$, which defines "closed in $A$". As $A \subseteq (X,\tau)$ is a subset of a topological space, it's natural to assume that it carries the (from $\tau$) induced topology $\tau_A$.

Now here we face the next difficulty: What is $\tau_A$? As I have mentioned yesterday, a set $B \subseteq A$ is said to be open in $\tau_A$, if and only if there is a open set $C \subseteq (X,\tau)$ of $X$, such that $B=C \cap A$. However, this is very close to the statement that we want to prove.

... if and only if there exists a closed subset $C \subset X$ such that $B = C \cap A$

So the question is: which definition of the topology $\tau_A$ of a subset $A$ do you actually use? It is always a good start to list what is given, i.e. which definitions, theorems and so on. It is the reason why we require part 2 of our template to be filled out. And in this case it is even more important, as the statement which has to be proven can as well be taken as the definition of $\tau_A$ and nothing would be left to show. So the only difference between the definition of $(A,\tau_A)$ and the statement is the fact, that one deals with open and the other one with closed sets? In this case we have to deal with the complements.

Let $B \subseteq (A,\tau_A)$ be a closed set, i.e. $A-B \subseteq (A,\tau_A)$ is open and by definition $A-B = C \cap A$ for a open set $C \subseteq (X,\tau)$. ...

Now you have to find a closed set $D \subseteq (X,\tau)$ for which $B=D \cap A$ to prove $"\Longrightarrow "$ and in a further step $"\Longleftarrow "$.

 

[sa1988]

In which case I best add the little bit in brackets that I thought wasn't vital to the question!

It should read:

"Let $A$ be a subspace of a topological space $X$ (that is, with the subspace topology), and let ... etc etc..." <-- note the part in brackets which I missed out in the original question wording. Sorry!

I'm a little confused by what you said. Why does it need this extra set $D$ ?

Is it not possible to use open neighbourhoods and show that they overlap on boundaries where they shouldn't?

Thanks.

 

[fresh_42]

I'm a little confused by what you said. Why does it need this extra set $D$ ?

I ran out of letters. You may switch $C$ and $D$ if you like or use $C_1$ and $C_2$. Since we only have the definition of "open", I used up $C$ for the open set $A-B=C \cap A$ and had to take a new letter for the set in the statement $B=D \cap A$ which is supposed to be closed. Of course, $D$ and $C$ are closely related, since $C$ is our only candidate to define $D$, but it cannot be equal, because on is open ($C$ in my notation) and one must be constructed closed ($D$ in my notation).

Is it not possible to use open neighbourhoods and show that they overlap on boundaries where they shouldn't?

Maybe, but then you bring in additional sets, namely boundaries (which you haven't defined, yet) and plenty of neighborhoods (one for each point in the boundary). I don't think this complication is necessary as we only have to switch closed to open and back. This can easily be done by only considering the complements. Just stay close at the definition, because that's all we have.

 

[sa1988]

I think it's this concept of complements which is killing me now. Everything seems to be continually flipping between open and closed (or 'clopen' in other cases!).

It's interesting to see that this problem can be overcome with just these concepts, however. Certainly shows the importance of openness and closedness - which I don't think I've properly grasped yet.

So if we're looking at the original definition and the scenario in the question, we have $A \subset (X, \tau)$ , with the subspace topology.

So $A$ is a subset equipped with a topology and is therefore open.

Then introduce $C \subset X$ which is defined as closed, and create $B = A \cap C$.

So $B$ is the set created by the intersection of $A \subset X$ (open) and $C \subset X$ (closed), and therefore $B$ is closed.

Now to show the 'only if' part:

Construct $B$ with the intersection of $A$ (open) and $D$ (open), then $B$ is open, because the intersection of any open sets it open.

Hence $B$ is closed if and only if $B = A \cap C$ where $C \subset X$ is closed.

This is honestly all I can fathom right now, and I'm pretty sure I'm still missing something.