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i need help with this exercise.

Prove $(-\infty,5]$ is not compact.

Edit: I prove is closed, and thanks for the users clear my confusion with this question. Thanks

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    It actually is closed, so I don't see what you're supposed to do. Maybe you need to show it is not compact? – 2017-01-22
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    It *is* a closed set in the usual topology on $\Bbb R$. – 2017-01-22
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    Perhaps you are trying to prove that it is not open? – 2017-01-22
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    Judging by the mention of Heine-Borel, I would guess that the objective is to show that it is not compact. – 2017-01-22
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    Yes @Ian, wait a moment i edit my quesiton – 2017-01-22
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    @copper.hat sorry i was confuse, i suppose that set is not closed ._. thanks. let me try to prove that – 2017-01-22
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    The set is closed.... ? – 2017-01-22
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    @Antonios-AlexandrosRobotis yeah. but your answer of compact is clear. thank – 2017-01-22

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This set is not compact, because consider the open cover consisting of $B(1,p)$ such that $p\in (-\infty,5]$. This cover can be shown to not reduce to a finite subcover. So, the set is not compact.

Note: here my notation for balls is $B(r,q)$, where $q$ is the center point, and $r$ is the radius.

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    I think you actually want the cover to be something like $B(5,p)$ where $p \in (0,\infty)$. – 2017-01-22
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    Why does the cover I proposed not work? – 2017-01-22
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    What does a ball of negative radius mean? (Or maybe our notation for balls is different...) – 2017-01-22
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    The balls are all of radius $1$, centered at points $p$. Sorry if this was unclear, I am using Folland's notation. – 2017-01-22
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    Ah, OK, yeah we just have the order of the arguments reversed. Carry on! (I usually write either $B(x,r)$ or $B_r(x)$.) – 2017-01-22
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    I'll note this! Thanks for bringing it up. – 2017-01-22
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Let $A_x = (x-1, x+1) \forall x \in \mathbb{Z}$ Then ${A_x}$ forms an open cover of $(-\infty, 5]$ for which every subcover is countably infinite.

Moreover, every $E \subset \mathbb{R}$ is compact iff it is closed and bounded, and your set is not bounded below.