# Open or closed subset respect to the Subset Topology

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 $$Y$$?

• 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. Oct 2, 2020 at 15:43
• @TheSilverDoe what about the definition of this topology? It says $U$ open in $X$, but in my example $A$ is not open in $\mathbb{R}$
• $A$ is not open in $\Bbb R$, right. But it is open in $Y$ endowed with the subspace topology Oct 2, 2020 at 16:01
• @InsideOut so the same argument holds for any interval in $\mathbb{R}$?
• 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? Oct 2, 2020 at 16:16
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.