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Are there any simple examples of subspaces of a normal space which are not normal?

I know closed subspace of a normal space is normal, but open subspace in most cases which I can think of are also normal.

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    $\begingroup$ There is a well known book Counterexamples in Topology that you can refer to for any standard counterexample. $\endgroup$ Oct 25, 2012 at 2:57
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    $\begingroup$ Why are you considering only open and closed subspaces? $\endgroup$ Dec 21, 2012 at 12:36
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    $\begingroup$ Some such examples can also be found in the database $\pi$-base: topology.jdabbs.com/theorems/I000036 topology.jdabbs.com/… $\endgroup$ Feb 12, 2019 at 17:33

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One well-known example is $X=(\omega_1+1)\times(\omega+1)$, where $\omega_1$ is the first uncountable ordinal with the order topology, and $\omega$ is the first infinite ordinal. $X$ is the product of two compact Hausdorff spaces, so $X$ is compact and Hausdorff and therefore normal, but $X\setminus\{\langle\omega_1,\omega\rangle\}$ is not normal: the closed sets $\omega_1\times\{\omega\}$ and $\{\omega_1\}\times\omega$ cannot be separated by disjoint open sets. $X$ is often called the Tikhonov plank, though the name is also sometimes applied to $X\setminus\{\langle\omega_1,\omega\rangle\}$.

Added: We can also appeal to the theorem that a space $X$ is Tikhonov (i.e., $T_1$ and completely regular) iff it is homeomorphic to a subspace of some product of closed unit intervals. Every product of closed unit intervals is compact and Hausdorff, hence normal, but there are many examples of Tikhonov spaces that are not normal, and every non-normal Tikhonov space is an example of a non-normal subspace of a normal space. A couple of the better-known examples of non-normal Tikhonov spaces are the Moore plane and the Sorgenfrey plane. Another example is the space described in this answer, if $\mathscr{D}$ is taken to have cardinality $2^\omega=\mathfrak c$, which the addendum to the answer shows to be possible; non-normality of the space follows from Jones’s lemma.

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    $\begingroup$ Sorry, am a newbie, and not familiar with "first uncountable ordinal", ω1, ω etc! $\endgroup$ Oct 25, 2012 at 1:20
  • $\begingroup$ @user73412: The space $\omega+1$ is homeomorphic to the subspace $\{0\}\cup\left\{\frac1n:n\in\Bbb Z^+\right\}$ of $\Bbb R$. Think of $\omega_1+1$ as a set $Y$ with a linear order $\le$ on it such that (a) every non-empty subset of $Y$ has a $\le$-least element; (b) $Y$ has a $\le$-greatest element called $\omega_1$; $Y$ is uncountable; and if $y\in Y\setminus\{\omega_1\}$, then $\{z\in Y:z\le y\}$ is countable. That won’t tell you how to work with it, but it may at least give you some idea of what it is. $\endgroup$ Oct 25, 2012 at 1:31
  • $\begingroup$ en.wikipedia.org/wiki/First_uncountable_ordinal $\endgroup$
    – arsmath
    Dec 17, 2013 at 15:47
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Consider $ [0,1]^{\mathbb{R}} $ with the product topology. By Tychonoff theorem $ [0,1]^{\mathbb{R}} $ is compact. Since every compact space is normal, we have $[0,1]^{\mathbb{R}} $ is normal.

Note that $ (0,1)^{\mathbb{R}} $ is a subpace of $ [0,1]^{\mathbb{R}} $. Since the product of uncountably many copies of ${\mathbb{R}}$ is not normal and ${\mathbb{R}}$ is homeomorphic to (0, 1) we have $ (0,1)^{\mathbb{R}} $ is not normal. Therefore a subspace of a normal space is not necessarily normal.

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    $\begingroup$ Did you mean "every compact Hausdorff space is normal"? $\endgroup$
    – Akira
    Oct 4, 2022 at 13:58
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    $\begingroup$ @Akira I think that’s what he meant lol $\endgroup$
    – homosapien
    Jan 17, 2023 at 16:51
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${\mathbb R}$ is homeomorphic to $(0,1)$ and Tikhonov theorem tells us that the product of countably infinitely many copies of an interval is compact Hausdorff so a subspace of a normal subspace is not always normal.

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  • $\begingroup$ I don't understand this example. What is the non-normal subspace? $\endgroup$
    – JSchlather
    Oct 25, 2012 at 2:00
  • $\begingroup$ Are you referring to my answer? $\endgroup$
    – glebovg
    Oct 25, 2012 at 2:20
  • $\begingroup$ Yes. I am referring to your answer. It seems to imply that $(0,1)$ is not normal. $\endgroup$
    – JSchlather
    Oct 25, 2012 at 2:35
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    $\begingroup$ ${[0,1]^{\mathbb R}}$ is normal but ${(0,1)^{\mathbb R}}$ is not. $\endgroup$
    – glebovg
    Oct 30, 2012 at 14:33
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    $\begingroup$ @glebovg That last comment ought to be part of the answer. Preferably with an explanation why $(0,1)^{\mathbb{R}}$ is not normal. As is, the answer evokes an "Uh, what?" reaction, it's not immediately clear where you are going. $\endgroup$ Jun 10, 2015 at 6:38
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Let, $X= \{a,b,c,d\}$ And $T= \{ \emptyset ,X,\{d\} , \{b,d\} ,\{ c,d\},\{b,c,d\}\}$ Then $(X,T)$ is a topological space. Since $(X,T)$ has no pair of disjoint non-empty closed sets, $(X,T)$ is a normal space.
Consider ,$Y=\{b,c,d\}$ of $X$. Then $T(Y)= \{\emptyset,Y, \{d\} , \{b,d\} ,\{ c,d\}\}$ Then $\{b\}$ and $\{c\}$ are disjoint closed sets in $(Y,T(Y))$ and they cannot be separated in $(Y,T(Y))$. Hence ,$(Y,T(Y))$ is not a normal space.

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The topology $T = \{\emptyset, X , \{a\} , \{a ,b\}, \{a , c\}, \{a ,b ,d\}\}$ on $X = \{a, b , c , d \}$ is normal. Take $Y = \{a, b, c\}$. Then the topology on the set Y is $S = \{ \emptyset , Y , \{a\} , \{ a ,b \} , \{ a , c \}\}$ is not normal. Hence subspace of normal space need not be normal.

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    $\begingroup$ Sorry, I think (X,T) itself is not normal. By definition, any normal space should be T1, and (X,T) is not T1. For example, pick a,d. There are no neighborhoods of a and d, disjoint from each other. $\endgroup$ Dec 31, 2012 at 10:30
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The topology $T=\{\emptyset , X,\{a\},\{a,b\},\{a,c\}\}$ on $X=\{a,b,c,d\}$.

Take the subspace $Y=\{a,b,c\}$ with Topology $S=\{\emptyset,Y,\{a\},\{a,b\},\{a,c\}\}$. Then Y is not normal but $(X,T)$ is normal.

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    $\begingroup$ $T$ is not a topology for $X$; $\{a,b\}$ is open, as is $\{a,c\}$, but $\{a,b\} \cup \{a,c\} = \{a,b,c\}$ is not. $\endgroup$
    – JSQuareD
    Jan 6, 2015 at 15:46

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