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I am reviewing some topics from point set topology from Munkres book and I would like to ask you the following question. The questions may sounds stupid but anyway I would be happy to clarify it:

Definition: Let $X$ be a topological space. Separation of $X$ is a pair of open, disjoint, nonempty sets $U,V$ such that $X=U\cup V$. We say that $X$ is connected space if there is no such separation of $X$.

Question: Let $X$ be a topological space and $A_{\alpha}\subset Y\subset X$. If $A_{\alpha}$ is connected subspace of $X$ then $A_{\alpha}$ is connected subspace of $Y$.

Can anyone explain to me is it true? And if yes how to understand that it is true. I would be very grateful for detailed help!

RFZ
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3 Answers3

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Let's say $\Omega_X$ is the topology on $A_\alpha$ which is induced from $X$ and $\Omega_Y$ is the topology on $A_\alpha$ which is induced from $Y$. Then $\Omega_X=\Omega_Y$. Indeed, let $T\in\Omega_X$. By definition there is an open set $U$ in $X$ such that $T=U\cap A_\alpha$. Since $A_\alpha\subseteq Y$ we have $T=U\cap A_\alpha=(U\cap Y)\cap A_\alpha$. Since the set $U\cap Y$ is open in $Y$ we conclude that $T\in\Omega_Y$.

For the other direction let $T\in\Omega_Y$. Then there is an open set $U$ in $Y$ such that $T=U\cap A_\alpha$. But since $U$ is open in $Y$ there is a set $V$ which is open in $X$ such that $U=V\cap Y$. Then $T=V\cap Y\cap A_\alpha=V\cap A_\alpha\in\Omega_X$.

So yes, if the topological space $(A_\alpha,\Omega_X)$ is connected then $(A_\alpha,\Omega_Y)$ is connected as well, since it is the same space.

Mark
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  • So when say that a subspace $Y$ of topological space $X$ is connected we mean its connectedness in subspace topology right? In other words, there is no separation for $Y$ with open sets in $Y$. – RFZ Sep 27 '19 at 15:52
  • Yes. And that's not just connectedness, but pretty much any topological property. We always think of a subspace with the induced topology. – Mark Sep 27 '19 at 15:53
  • Since $\Omega_X=\Omega_Y$ then if $(A_{\alpha},\Omega_Y)$ was not connected then exists separation of $A_{\alpha}$ into open sets from $\Omega_Y$ but from equality of topologies it will lead that it is not connected in $(A_{\alpha},\Omega_X)$, which is wrong. RIght? – RFZ Sep 27 '19 at 15:56
  • There are of course properties like "dense in" that talk about a space and a subset. Connectedness is not one of them. – Carsten S Sep 27 '19 at 16:05
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    @ZFR $(A_\alpha,\Omega_X)$ and $(A_\alpha,\Omega_Y)$ is exactly the same space. So if you get a separation in one of them then of course this is also a separation in the other space. – Mark Sep 27 '19 at 16:48
  • Dear Mark! Thanks a lot for your help! I appreciate it! – RFZ Sep 27 '19 at 16:56
  • Dear Mark! Could you take a look at my last topic please? https://math.stackexchange.com/questions/3372643/linear-continuum-in-order-topology-is-connected – RFZ Sep 28 '19 at 00:52
  • I only saw the question now. Well, as I see you already got a few answers there anyway. – Mark Sep 28 '19 at 08:55
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You did not define a "connected subspace" but only a "connected space". So if $X$ is a topological space and $A\subset X$ then, by definition, $A$ is connected iff it is connected as a topological space, endowed with the subspace topology. So your only question here is: If $A\subset X\subset Y$, $Y$ a topological space, $X$ endowed with the sub-space topology wrt $Y$, is the sub-space topology on $A$ as a subset of $X$ the same as the sub-space topology on $A$ as a subset of $Y$? The answer is "yes", and I leave the proof to you. (Of course, if the topologies on $X$ and $Y$ were unrelated, the answer would be "no" in general.)

Carsten S
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The key is that the sets are open with respect to the subspace topology of $A_\alpha$

Reveillark
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  • Could you explain it in more rigorous way? I know the subspace topology but i cannot relate all this. Probably that's because i am learning topology for short time. – RFZ Sep 27 '19 at 15:35
  • But $A_{\alpha}$ can be regarded as a subspace of $Y$ and $X$. Anyway could you give more details? – RFZ Sep 27 '19 at 15:37
  • For $U$ to be open in the subspace topology of $A$ in the ambient space $Z$, $U$ should be of the form $A\cap U_1$ for some $U_1$ open in $Z$. What does this mean when $A=A_alpha$ and $Z=X$? What about $A=A_alpha$ and $Z=Y$? And if $A=Y$ and =Z=X$? – Reveillark Sep 27 '19 at 15:49