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I was doing a problem based on constructing an interval centered on $c$ which is a subset of a separate interval $(a,b)$ , where $c \in (a,b)$ and $a$ and $b$ are real numbers.

The issue with trying to create an interval like $I = (c-(c-a), c+(c-a))$ is that $c$ may be closer to $b$ than to $a$, and thus adding the difference between $c$ and $a$ would exceed $b$, and $I$ wouldn't be a subset of $(a,b)$.

I was looking at the solution manual and it said: Let $r=\min (c-a, b-c)$, and let $I = (c-r, c+r)$.

From this it seems that $\min$ means select $r$ to be the smaller of the two numbers in the parentheses? I've never seen this operation before and it's not in the textbook. Is it just shorthand for :

Let $r$ be an element of the real numbers : if $(c-a)>(b-c)$ then $r = b-c$ . If $(c-a)<(b-c)$ then $r=c-a$. If $(c-a)=(b-c)$ then $r=c-a=b-c$ ?

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3 Answers

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Yes, minimum of the two numbers in the parentheses.

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Note that the minimum is not always well-defined. For example, given the interval $(0,1)$, there is no smallest element in the interval, since we may find numbers in the interval arbitrarily close to zero. So the minimum (and maximum) are not always well defined. We may use infimums and supremums to fix this problem.

When learning about the min and max operations, I think it's important to know their limitations.

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"Min" means minimum, so in your case it's saying what's the minimum of the two values in (c-a, b-c).

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