Is there a largest open interval for an open set (not necessarily bounded)?

If $$G$$ is an open subset of $$R$$, and if $$x\in G$$, show that there exists a largest open interval $$I_x$$ containing $$x$$ s.t $$I_x$$ is the subset of $$G$$.

My idea:

Let $$x\in (a_x,b_x)$$ where
$$a_x=\inf\{a and
$$b_x=\sup\{b>x|(x,b)\subset G \}$$.

Let $$I_x=(a_x,b_x)$$.

I want to show $$a_x$$, $$b_x$$ can not belong to G, hence $$I_x$$ is the largest interval.

Assume $$a_x\in G$$, this contradicts the fact that $$a_x$$ was $$\inf$$. so $$a_x$$ is not in $$G$$. Likewise for $$b_x$$.

I think if it was said that $$G$$ is bounded, I could confidently use the proof idea above. But it is NOT. So what if G is unbounded? Then I may not have finite $$a_x$$ and $$b_x$$. Or do I need to be worried about this at all?

You could change your approach to defining $$a_x$$ and $$b_x$$ ever so slightly, to make your proof correct.

• If the set $$\{a is bounded below, then let $$a_x$$ be the infimum of that set. Otherwise let $$a_x = -\infty$$.
• If the set $$\{b>x \mid (x,b)\subset G \}$$ is bounded above, then let $$b_x$$ be the supremum of that set. Otherwise let $$b_x = +\infty$$.

Now you should go through the remainder of your proof and check carefully for any changes that are required by having changed the definitions of $$a_x$$ and $$b_x$$.

• Thank you very much. – BesMath Mar 14 at 16:39
• Not that I disagree with this solution, but just in order to not mislead the OP, in this case $a_x$ and $b_x$, when they are equal to $\infty$ or $-\infty$, are not real values anymore. Indeed they become just symbols helping to define the interval. – almaus Mar 14 at 16:39

Every open set in $$\mathbf{R}$$ can be written as a countable union of pairwise disjoint open intervals. So we get $$G=\bigcup_{i=1}^{\infty}I_i$$ Let $$x \in G \implies \exists! n\in\mathbf{N}$$ so that $$x\in I_n$$

Now you can check that this interval will be your maximal interval containing $$x$$ that's contained in $$G$$.

• You are using a bazooka theorem here... harder to show that than the original problem. – almaus Mar 14 at 16:18
• @almaus I agree but it's good to see different ways to prove things – guy3141 Mar 14 at 16:19

You can just say that as $$G$$ is open, it means that there exists an open interval included in $$G$$ around each of its points, thus there exists at least $$a_0$$ and $$b_0$$ such that $$x \in (a_0, b_0) \subset G$$.

And then you can consider the set of all the intervals in $$G$$ including $$x$$, and take the biggest.

• That is exactly where I have a problem with. since G is open for each element x in G, there are point less and greater that x, so no sup or inf has a place of meaning here. so how to make the largest interval? I am really confused. – BesMath Mar 14 at 16:27
• You need not be afraid of infinites here, as $(b, \infty)$ or $(-\infty, a)$ are perfectly valid open intervals of $\mathbb{R}$. – almaus Mar 14 at 16:29
• So just make two cases (for each boundary): when there exists an upper (resp. lower) bound, and when there is not. In the case there is not, that just means that your upper (resp. lower) boundary is $\infty$ (resp $-\infty$). – almaus Mar 14 at 16:32

Your proof is fine when situated in the appropriate framework.

Although it might not be accepted in the context of a class, I think the best way to handle this problem is to use the extended real line $$\bar{\mathbb R} = \mathbb R \cup \{-\infty,\infty\}$$ where we can order this set in the obvious way and define infimum and supremum from the order - and likewise, can define open intervals as usual, with the observation that $$(-\infty,x)$$ are honest intervals in this view that coincide with the usual definitions. The importance of this change is that every set has a supremum and infimum in the extended reals - so you do not need to worry about boundedness at all.

Basically, with this change of context, you just say that you have some subset of $$\mathbb R$$ and let $$a_x$$ and $$b_x$$ be the infimum and supremum of that set in $$\bar{\mathbb R}$$, and then just finish your argument exactly as you did - except that you might, for completeness, observe that if $$a_x$$ and $$b_x$$ are real, they are not in the set for the reasons you observe, and if they are not, they are not in the set because the set is a subset of $$\mathbb R$$.

One often finds that analysis questions such as this one are much clearer if you work with $$\pm \infty$$ within the domain of mathematics, rather than, as is common, saying that every expression involving $$\infty$$ is specially defined and requires casework - because often the extended reals unify the theory with no need for extra work.