# Homework Help: Topology by Simmons Problem 1.2.1

Tags:
1. Feb 24, 2017

### Figaro

1. The problem statement, all variables and given/known data
If $\bf{A}$ $= \{A_i\}$ and $\bf{B}$ $= \{B_j\}$ are two classes of sets such that $\bf{A} \subseteq \bf{B}$, show that $\cap_j B_j \subseteq \cap_i A_i$ and $\cup_i A_i \subseteq \cup_j B_j$

2. Relevant equations

3. The attempt at a solution
Since $\bf{A} \subseteq \bf{B}$, for every $A_i$ such that for all $x\in A_i$ there exist a $B_j$ such that $x\in B_j$

For $\cap_j B_j \subseteq \cap_i A_i$,

For every $x\in \cap_j B_j$ and since all $x$ is also in $A_i$, then $x\in \cap_i A_i$. Thus $\cap_j B_j \subseteq \cap_i A_i$

For $\cup_i A_i \subseteq \cup_j B_j$,

Since $\bf{A}$ $= \cup_i A_i$, $\bf{B}$ $= \cup_j B_j$, and from $\bf{A} \subseteq \bf{B}$. Thus, $\cup_i A_i \subseteq \cup_j B_j$

2. Feb 25, 2017

### Staff: Mentor

Why are all $x$ in $A_i$? $\bf{B}$ has more sets than $\bf{A}$, so there might be elements in some $B_j$ which are not contained in any $A_i$. What's with them?
$\bf{A} \neq \cup_i A_i$ and $\bf{B} \neq \cup_j B_j$
If this were the case, then nothing would have to be shown here.

Hint: Define a function $\{i\} \rightarrow \{j\}$ and consider in the first case a fixed, but arbitrary element $x$ of $\cap_j B_j$ and in the second case a fixed, but arbitrary element $x$ of $\cup_i A_i$. What does it mean, that these choices belong into these sets?
By the way, do you recognize, that it's a disadvantage not to have named the index sets?

3. Feb 25, 2017

### Figaro

Yes, I made a mistake in your first statement, but why is $\bf{A} \neq \cup_i A_i$ and $\bf{B} \neq \cup_j B_j$

What do you mean by defining a function $\{i\} \rightarrow \{j\}$? I know that $i\leq j$. For a fixed but arbitrary $x$ of $\cap_j B_j$, $x$ should belong to all $B_j$ and in effect it would belong to at least some $A_i$ since for some $x$ in $A_i$ it also belongs to $B$. Thus $\cap_j B_j \subseteq \cap_i A_i$.

For the second, for some $x$ in $\cup_i A_i$, this means that x belongs to at least some $A_i$ but by hypothesis $\bf{A} \subseteq \bf{B}$, all sets $A_i$ belongs in $\bf{B}$ so for sure that $x$ also belongs to some $B_i$. Thus $\cup_i A_i \subseteq \cup_j B_j$

4. Feb 25, 2017

### Staff: Mentor

No. $i \leq j$ doesn't make sense and is a direct consequence of ignoring the index sets.

As always, a list of evidences is helpful. We have two sets of sets or classes, but this isn't important here.
$\mathcal{A}=\{A_i\,\vert \,i\in \mathcal{I}\}$ and $\mathcal{B}=\{B_j\,\vert \,j\in \mathcal{J}\}$.
Then we have $\mathcal{A}\subseteq \mathcal{B}$.

So first of all, where do you get, e.g. $\mathcal{A}=\bigcup_{i\in \mathcal{I}}A_i$ from? $\mathcal{A}$ is a set (class) of sets, not their union.
Now what does this inclusion $\mathcal{A}\subseteq \mathcal{B}$ mean? It means, that every element of $\mathcal{A}$, that is every set $A_i$ is an element of $\mathcal{B}$, i.e. a set $B_j$. But this $j$ depends on $i$, so we have a function $\tau \, : \,\mathcal{I} \longrightarrow \mathcal{J}$ and an equation $A_i = B_j=B_{j(i)}=B_{\tau(i)}$.

This is the complete list of given conditions. Nothing specially done so far. I only put more effort in describing the exact situation, as it is always a good habit to start with. It makes everything next simpler and we have a common base to talk, which is by the way the real reason for the points 1 and 2 of our template. It helps to avoid misunderstandings. It is also a good advice for exams. It might look like more time and trouble at the beginning, but it saves more time than it costs during the solution and often helps to avoid mistakes.
No. "At least" doesn't help us, as we want to show, that $x$ belongs to all $A_i$ and not just one.

So let's have this $x \in \bigcap_{j\in \mathcal{J}} B_j$. Now we know, that some $B_{\tau(i)}$ equal a $A_i$ and some $B_j$ do not. Does it matter, if they don't? Do the $B_{\tau(i)}=A_i$ cover all $A_i$? Can you go on from here?

5. Feb 25, 2017

### Figaro

I think I get what you are saying, it doesn't matter if some $B_j$ are not equal to some $A_i$ since we are getting the intersection.
Ok, so for all $x$ in $\cap B_j$, and due to the fact that all $A_i$ are equal to some $B_j$, all $A_i$ "contribute" to the intersection, but this implies that for every $x$ in $\cap B_j$, $x$ is also in $\cap A_i$. Thus $\cap B_j \subseteq \cap A_i$.

Is my proof for the union correct?

6. Feb 25, 2017

### Staff: Mentor

Yes, that's correct, and if you substitute "contribute to" by "are parts of" it's perfect.

You could also say $\bigcap_{j\in \mathcal{J}}B_j \subseteq \bigcap_{i\in \mathcal{I}}B_{\tau (i)}=\bigcap_{i \in \mathcal{I}}A_i$ because the latter intersection is over fewer sets.