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Could somebody please help me clarify this statement?

Let $X=\mathbb{R}$ with the usual topology, $E=[a,b]$. Then $[a,x)$ is open in the induced topology for every $x\in[a,b]$.

What does "open in the induced topology" mean?

I tried to look elsewhere but could not find the exact definition of "open in the induced topology". Could somebody please give some light on this?

How can we explain using the concept of "open" as in open set? Or is there any other meaning to it?

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It's the subspace topology. The open sets of $E$ are sets of the form $U\cap E$, where $U$ is an element of $X$'s topology, ie, $U$ is open in $X$.

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  • $\begingroup$ But the element of $X$'s topology should be of the form $(x,y)$ isn't it? But $[a,x)$ is not in that form. Could you clarify on this? Thanks. $\endgroup$ – user338393 Sep 19 '16 at 6:29
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    $\begingroup$ @user338393 but $[a,x)=E\cap (a-1,x)$ where $(a-1,x)$ is open in $X$ $\endgroup$ – Hagen von Eitzen Sep 19 '16 at 6:31
  • $\begingroup$ @user338393 what you are thinking of would be considering the open sets $U$ of $X$ that are subsets of $E$ but this is generally not the same. As shown by Hagen's example. $\endgroup$ – quid Sep 19 '16 at 6:36

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