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More specifically, let $O_1, . . . , O_n$ be a finite collection of open subsets of the continuum, $C$. Then the intersection $O_1 ∩ · · · ∩ O_n$ is open as well. I think it is possible to do it without considering metric spaces (and therefore balls of radius $r$). Also, if you could prove why the intersect of an infinite amount of open subsets turns out to be possibly closed, it would be very much appreciated as this is difficult to wrap my head around.

Also open sets are defined as sets that don't contain their endpoints. Sorry for being unclear.

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How do you define your open sets without metric spaces? If you are working in a a topological space doesn't what you want follow from the definition. –  Shahab Sep 20 '12 at 5:14
    
I edited the question. Thanks for the suggestion. –  Casquibaldo Sep 20 '12 at 5:33
    
I think you need to recheck your definition of open sets. What is the meaning of end point? Both the set of rational numbers and the set of integers will be open sets as per your definition. –  Shahab Sep 20 '12 at 6:02
    
I should have phrased open sets as open is their complement was closed. Sorry. –  Casquibaldo Oct 3 '12 at 16:50
    
Perhaps you should define open sets to be sets which are countable unions of open intervals. –  Shahab Oct 4 '12 at 3:18

4 Answers 4

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A set $A$ is open if every $x \in A$, there exists some $\epsilon > 0$ such that $B(x,\epsilon) \equiv \{y : y \in C,\, d(x,y)<\epsilon \} \subset A$ i.e. $x$ is in some open ball that is in $A$, i.e. every $x \in A$ is an interior point.

Let $X = O_1 \bigcap O_2$. Then any $x \in X$ implies $x \in O_1$ and $x \in O_2$, implying that you should always have an open ball around $x$.

We prove via contradiction. Now assume $X$ is not open. Since $X$ is not open, there exists $x \in X$ such that $x$ is on the boundary of $X$. This implies that $x$ is on the boundary of either $O_1$ or $O_2$. But this is a contradiction, since $x$ has to be an interior point of both $O_1$ and $O_2$. Therefore, all elements of $X$ are interior points. Therefore, $X$ must be open.

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For your second question, for any positive integer $n$, let $A_n=(-1/n,1/n)$.

The intersection of all the $A_n$ is $\{0\}$.

Or else let $B_n=(0,1/n)$. The intersection of all the $B_n$ is the empty set, which is closed (and open).

The answer to the first question depends on the details of your definition of open. Let us define a set $A$ of reals to be open if for any $x\in A$, there is a positive $\epsilon$ (which usually depends on $x$) such that the interval $(x-\epsilon,x+\epsilon)$ is a subset of $A$.

Now let $x$ be in the intersection of your $O_i$. Then for every $i$, there is a positive $\epsilon_i$ such that $(x-\epsilon_i,x+\epsilon_i)$ is a subset of $O_i$. Let $\epsilon$ be the smallest of the $\epsilon_i$. Then the interval $(x-\epsilon,x+\epsilon)$ is a subset of $O_1\cap O_2\cap\cdots\cap O_n$.

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If you're using a topological space, by definition the intersection of any two open sets gives an open set, so by repeating that finitely many times you end with an open set. For example, let $O_1$, $O_2$, and $O_3$ be open sets. $O_1 \cap O_2$ is open, so $(O_1\cap O_2)\cap O_3$ is also open. If there is also an $O_4$, then $\left(\left(O_1\cap O_2\right)\cap O_3\right)\cap O_4$ is open. This process can be carried on for any finite collection of open sets.

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Since the real line is a linearly ordered topological space, open sets may have been defined as those sets that are unions of open intervals, i.e., unions of sets of the form $(a,b)$. In that case suppose that $U_0$ and $U_1$ are open sets. Then there are index sets $A_0$ and $A_1$ and families $\mathscr{I}_0=\{I_{0,\alpha}:\alpha\in A_0\}$ and $\mathscr{I}_1=\{I_{1,\alpha}:\alpha\in A_1\}$ of open intervals such that $$U_0=\bigcup_{\alpha\in A_0}I_{0,\alpha}\qquad\text{and}\qquad U_1=\bigcup_{\alpha\in A_1}I_{1,\alpha}\;.$$

Then

$$\begin{align*} U_0\cap U_1&=\left(\bigcup_{\alpha\in A_0}I_{0,\alpha}\right)\cap\left(\bigcup_{\beta\in A_1}I_{1,\beta}\right)\\ &=\bigcup_{\alpha\in A_0}\left(I_{0,\alpha}\cap\bigcup_{\beta\in A_1}I_{1,\beta}\right)\\ &=\bigcup_{\alpha\in A_0}\left(\bigcup_{\beta\in A_1}\Big(I_{0,\alpha}\cap I_{1,\beta}\Big)\right)\\ &=\bigcup_{\langle\alpha,\beta\rangle\in A_0\times A_1}\Big(I_{0,\alpha}\cap I_{1,\beta}\Big)\;.\tag{1} \end{align*}$$

The intersection of two open intervals is an open interval (possibly empty), so $(1)$ is a union of open intervals and as such is an open set.

This shows that the intersection of two open sets is open, and the result for any finite collection of open sets is easily established by induction. Suppose that you know that the intersection of $n$ open sets is always open. Let $U_1,\dots,U_{n+1}$ be any $n+1$ open sets. Let $U=U_1\cap\ldots\cap U_n$; by the induction hypothesis $U$ is open. Then &U_1\cap U_2\cap\ldots\cap U_n\cap U_{n+1}=U\cap U_{n+1}$, which, being the intersection of two open sets, is open by the result proved above.

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