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Let $X$ be a topological space and let $Y$ be a subset of $X$. Observe that

$\{Y\cap U|U$ open set of $X$} is a topology on $Y$ called the subspace topology.

Let $X=\mathbb{R}$ and $Y =[0,1]\cup(2,3)$. Is the set $[0,1]$ open or closed as subspace of $Y$?

My attempt of the solution would be the following:

Let $A:=[0,1]$

Then with the subspace $\big(\frac{-1}{2},\frac32\big)$ we have $\big(\frac{-1}{2},\frac32\big)\cap Y=A.$ The subspace is open in $\mathbb{R}$, thus $A$ is open in $Y$.

On the other hand with the subspace $A$, which is closed in $\mathbb{R}$ we have $A\cap Y=A$. Hence $A$ is also closed in $Y$.

Is it correct to assume that $A$ is both open and closed in $Y$?

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    $\begingroup$ Yes it is. Actually this proves that $Y$ is not connected, since you found a closed and open subset of $Y$ which is not $\emptyset$ neither $Y$ itself. $\endgroup$ Oct 2, 2020 at 15:43
  • $\begingroup$ @TheSilverDoe what about the definition of this topology? It says $U$ open in $X$, but in my example $A$ is not open in $\mathbb{R}$ $\endgroup$
    – Dada
    Oct 2, 2020 at 15:49
  • $\begingroup$ $A$ is not open in $\Bbb R$, right. But it is open in $Y$ endowed with the subspace topology $\endgroup$
    – InsideOut
    Oct 2, 2020 at 16:01
  • $\begingroup$ @InsideOut so the same argument holds for any interval in $\mathbb{R}$? $\endgroup$
    – Dada
    Oct 2, 2020 at 16:08
  • $\begingroup$ I am not sure I really get what you mean.. given any open interval in $\Bbb R$, say $(a,b)$, then the intersection $(a,b)\cap\, Y$ is open in $Y$, by definition. Generally if $U\subset Y$ is open in $X$ then it is also open in $Y$. The converse is not true, an example given by your exercise above, $A$ is open in $Y$ but it is not in $X=\Bbb R$. Does it answer to your question? $\endgroup$
    – InsideOut
    Oct 2, 2020 at 16:16

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Yes, you're right: both $[0,1]$ and $(2,3)$ are open and closed (also called clopen) in $Y$. It shows that $Y$ is disconnected.

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