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Suppose we have this interval $A=]1, 5].$ We know that $A$ is in the closure of $A$.

If we add an arbitrary number $b > 0$ to $1$ we get: $1 + b > 1$, so the intersection of the sphere around $1$ with radius $b$ with $A$ is non-empty. So $1$ is in the closure of $A$, and the closure of $A$ is $[1,5]$.

Do I need to prove that that are no other terms in the closure of $A$? If so, how can I do this?

D.G. van de Schepop
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  • i would use the word "arbitrary" in stead of "random" – 311411 Apr 06 '22 at 14:06
  • I edited it, thanks – D.G. van de Schepop Apr 06 '22 at 14:10
  • I guess you are dealing with the closure of $A$ as a subset of $\mathbb R$ where $\mathbb R$ is equipped with its usual topology. That is essential info and should be mentioned in your question. If e.g. you are looking for the closure of $A$ as a subset of $(1,\infty)$ or $(1,5]$ then things are different. – drhab Apr 06 '22 at 14:32

2 Answers2

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Since you appear to be at the level of elementary analysis, it seems a good idea to prove it. It is not very difficult. You already understand about the key ideas of "arbitrarily small positive number" $b$, and then using $b$ as a radius of a little ball.

If you pick an arbitrary real $t$ such that $t<1$ or $t>5$, you can prove that $t$ is not in the closure of $A$. This you can do by specifying a radius $b=b(t)$ so that the sphere around $t$ of radius $b$ satisfies $\_\_\_\_(??)\_\_\_\_.$

Can you say what the missing condition (??) must be? If not, try the specific value $t=7$, and draw a little picture.

311411
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Take any point $x \not\in [1,5]$, and show that $x$ is not a limit point of $A$. Suppose $x < 1$ (the case $x > 5$ is symmetric).

Now let $\epsilon = \dfrac{1-x}{2}$ and you have found an open neighbourhood $(x-\epsilon, x+\epsilon)$ which does not intersects with $A$, which by definition implies that $x$ is not a limit point of $A$.

Therefore, you found all limit points of $A$ and $[1,5]$ is its closure.

ObliviousCapybara
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