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I hold two basic opinion that

(1) Every line segment has two endpoints.

(2) The endpoint of a line segment is part of it.

So as for the line segment corresponding to [0,1], if we remove its right endpoint, which number corresponds to the right endpoint of the new line segment? I don't think the number is still 1, because we have already removed the point corresponding to 1, thus created this new line segment, the point 1 is not part of the new line segment, so according to my opinion (2), I don't think the number is 1.


Update: A bounded continuous segment in a line is my definition of "line segment"


Update: I think I solved my question , see answer below

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    $\begingroup$ What is your definition of "line segment"? I would say that $[0,1)$ isn't a line segment at all. $\endgroup$ Dec 28, 2020 at 0:04
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    $\begingroup$ Well then what do you mean by "segment"? $\endgroup$ Dec 28, 2020 at 0:09
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    $\begingroup$ The right end point is still $1$ and it is a half-open line segment. A line segment may not contain its end points -- it only needs to contain all points between them. Read here $\endgroup$ Dec 28, 2020 at 0:10
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    $\begingroup$ The interval $[0,1]$ has an infimum $0$ which happens to also be a minimum and also a supremum $1$ which also happens to be a maximum. The half-open interval $[0,1)$ has an infimum $0$ which happens to also be a minimum and also a supremum $1$ which is not a maximum since it is not included in the half-open interval. The half-open interval $[0,1)$ does not have a maximum. "(2) The end-point of a line segment is part of it" However you make your definitions you are using rigorous, this statement will likely be false. $\endgroup$
    – JMoravitz
    Dec 28, 2020 at 0:15
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    $\begingroup$ Now.... you might be tempted to say that after removing $1$, there is still going to be some "largest" number in the interval... that it would be something like $0.999\dots$, however recall that $0.999\dots$ is identically equal to $1$. With that in mind, any proposed maximum $m$ which is strictly less than $1$ we can show could not be the maximum since looking at the average, $\frac{m+1}{2}$, this will be strictly greater than $m$ and strictly less than $1$ and hence a number larger than $m$ in the interval contradicting that $m$ could have been the maximum in the first place.. $\endgroup$
    – JMoravitz
    Dec 28, 2020 at 0:22

4 Answers 4

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Your opinions (1) and (2) are inconsistent with your (implied) opinion (3), that $[0,1)$ is a line segment. If you truly believe all three of these statements, then it follows, as night follows day, that you believe that $2+2=7$, that the Moon is made of green cheese, and that you are the Pope. I would recommend that you carefully consider your three beliefs, and see which one(s) you are really committed to, and which one(s) you are willing to abandon, in the interests of not being in a state of self-contradiction.

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Your second "basic opinion" is simply untrue. The number is $1$.

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There strictly speaking is no endpoint of a line_segment or ray at an open boundary. That said, the real number corresponding to an end of a line_segment defined by and terminating at a finite value {be it [included or excluded)} is simply that number, i.e. 1.

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First , it is helpful to inform me to check whether [0,1) has a right endpoint after removing .

Second , just as @ShubhamJohri informed that

A line segment may not contain its end points -- it only needs to contain all points between them. Read here

so I misunderstood the definition of a line segment.

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