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Is interval $(a, \infty]$ a Borel set ?

I hear that if $U$ is written as union, intersection, or subtraction of open sets, $U$ is called a Borel set.

I wonder whether $(a, \infty]$ is a Borel set or not.

$\cup_{n=1}^{\infty} (a, n]$ isn't equal to $(a, \infty]$, and I tried to write $(a, \infty]$ as union, intersection, or subtraction of open sets but I couldn't.

How can I write $(a, \infty]$, or isn't $(a, \infty]$ a Borel set ?

Asaf Karagila
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daㅤ
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    What is the space you are working in? – Rushabh Mehta Jun 21 '21 at 15:03
  • $\newcommand{c}{\complement}$The usual definition states "relative component" rather than subtraction of open sets, but then again, $A\setminus B=A\cap B^\c$, so it is an equivalent definition. As already mentioned by others, the notion of a Borel set depends on the topological space you're working in, ie, a set $U$ is a Borel set in a topological space $X$ iff it can be represented from the open sets in $X$ using countable unions/intersections/relative complement. – Prasun Biswas Jun 21 '21 at 15:14
  • The family of Borel sets of a topological space $X$ is defined as the $\subset$-smallest $\sigma$-algebra on $X$ that contains all the open sets. – DanielWainfleet Jun 21 '21 at 21:56

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In which space are you working? If you are working in the extended real line with the order topology, then $(a, \infty]$ is open by definition of order topology, hence a Borel set.

Son Gohan
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