# Finite Intersection of Open Sets Are Always Open?

1. Dec 26, 2011

### gwsinger

Suppose we have non-empty $A_{1}$ and non-empty $A_{2}$ which are both open. By "open" I mean all points of $A_{1}$ and $A_{2}$ are internal points. There is an argument -- which I have seen online and in textbooks -- that $A_{1} \cap A_{2} = A$ is open (assuming $A$ is non-empty) since:

1. For some $x \in A, A_{1}, A_{2}$ we have neighborhoods $N_{1}$ and $N_{2}$ around $x$ of radii $r_{1}$ and $r_{2}$ respectively, such that $N_{1} \subset A_{1}$ and $N_{2} \subset A_{2}$. We know this since $A_{1}$ and $A_{2}$ are stipulated to be open.

2. If we take $min(r_{1},r_{2}) = r$ we can then construct a neighborhood $N$ of radius $r$ around $x$ and (this is the part that seems false to me) we can somehow KNOW that $N \subset A$ and therefore conclude that $A$ is open.

But why (2)? How do we KNOW that $N \subset A$? Why can't $N_{1} \subset A_{1}$ and $N_{2} \subset A_{2}$ but nevertheless $N \not \subset A$?

2. Dec 26, 2011

### gb7nash

After we have constructed N, we no longer care about N1 and N2 separately. We only care about N. The main reason is that we chose the smallest neighborhood. Suppose N = N1. Since this is the smallest neighborhood, we have:

$N \subset A_1$ and $N \subset A_2$, so... The other case is analogous to this.

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As a side note (and I think this is what you're talking about), N1 may not be a subset of A2. This is the reason why we want to choose the minimum radius from N1 and N2. If N1 is not a subset of A2, we make the new radius small enough so that it is a subset of A2. From the beginning, N2 is a subset of A2, so we'll just use N2's radius.

Last edited: Dec 26, 2011
3. Dec 26, 2011

### Fredrik

Staff Emeritus
You defined N as the smaller of N1 and N2. (Clearly one of them is a subset of the other). If $N=N_1$, then $N=N_1\subset A_1$ and $N=N_1\subset N_2\subset A_2$. This clearly implies that $N\subset A_1\cap A_2$.

4. Dec 26, 2011

### gwsinger

I'm still not getting it! Sure, $N$ is defined as the smaller of $N_{1}$ and $N_{2}$. So what? Why does that guarantee it will fit in $X$? Here is a visual aid to what I am sensing as a counter-example (see attached). In the picture we have $N_{1} \subset A_{1}$ and $N_{2} \subset A_{2}$ but clearly not $N_{1} \lor N_{2} \subset A$.

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Last edited: Dec 26, 2011
5. Dec 26, 2011

### lugita15

But N1 and N2 have to be centered at x. In other words, N1 consists of all the points within a distance r1 of x, and N2 consists of all points within a distance r2 of x. Either r1 is less than or equal to r2 or r2 is less than or equal to r1. So either all the points within a distance r1 of x are within a distance r2 of x, or vice versa. Thus either N1 is a subset of N2, or vice versa.

6. Dec 26, 2011

### Fredrik

Staff Emeritus
Yes, my answer is the same as lugita15's. I assumed that you (gwsinger) define "neighborhood around x"="open ball around x"="open ball with x at the center". If you define "neighborhood around x"="open ball that contains x", then you're right that the argument needs to be changed.

7. Dec 26, 2011

### gwsinger

I see your point -- but I thought the goal was to think outside of the R2 metric. In Principles of Mathematical Analysis 2.24, Rudin makes a proof along these lines but NOT just for R2. So the question becomes: is there something that makes the proof necessarily true even in obscure metrics? If not, I'm willing to bite the bullet on this one and accept that the intuition we gain from why this must be true in R2 somehow applies to other metrics as well.

Last edited: Dec 26, 2011
8. Dec 26, 2011

### lugita15

Nothing in my (or Fredrik's) post depends on having $R^{2}$ as the metric space. All I said is that if r1 is less than or equal to r2, then the points within a distance r1 of x are automatically within a distance r2 of x. In other words, if d(x,y)<r1 and r1≤r2, then d(x,y)≤r2.

9. Dec 26, 2011

### Fredrik

Staff Emeritus
This is how I'd do the proof for an arbitrary metric space X. For all $x\in X$ and all r>0, $B(x,r)$ denotes the open ball around x with radius r.

If $A_1\cap A_2=\emptyset$, then $A_1\cap A_2$ is obviously open. So suppose that $A_1\cap A_2\neq\emptyset$. Let $x\in A_1\cap A_2$ be arbitrary. Since $A_1$ is open, there's an $r_1>0$ such that $B(x,r_1)\subset A_1$. Since $A_2$ is open, there's an $r_2>0$ such that $B(x,r_2)\subset A_2$. Define $r=\min\{r_1,r_2\}$. Clearly, $B(x,r)\subset B(x,r_1)\subset A_1$ and $B(x,r)\subset B(x,r_2)\subset A_2$. So every member of $B(x,r)$ is a member of both $A_1$ and $A_2$. This means that $B(x,r)\subset A_1\cap A_2$.