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Let $(X,\tau)$ be a topological space satisfying $T_i$ axiom, and $\emptyset\neq Y \subsetneq X$ endowed with the induced topology via $\tau$ denoted by $\tau_Y$.

Is $(Y,\tau_Y)$ also a $T_i$-space?

As exerices for the class I enrolled this semester, I proved them up to $T_3$, and after checking some articles about these axioms, I figured out that all of these axioms force stronger sense of topologically distinguishability upon the space, in other words being distinct(not the same) points or subsets is not enough, we want stronger form of separability than of either distinctness(for points and subsets) or disjointness(for subsets). Thus, one can argue that this restriction can clearly hold in the subspace, because the subspace is its superspace except some points are absent. Is this kind of reasoning sufficient and rigorous? I'm a little insecure about it, but if I'd heard it from an instructor I would immediately add it to the system of my true statements.

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    $T_4$ isn't inherited by subspaces in general. – 2017-01-04
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    the definition? – 2017-01-04
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    Oh, you're right! $T_4$ needs subspaces to be a closed subset! @DanielFischer – 2017-01-04
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    A completely regular Hausdorff space can be embedded in a compact Hausdorff space (hence normal) whether or not it is normal. – 2017-01-04

1 Answers 1

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(1). A subspace $Y$ of a $T_{3\frac {1}{2}}$ space $X$ is also $T_{3\frac {1}{2}}.$

Let $p\in Y$ and $p\not \in S\subset Y$ , with $S=Cl_Y(S)=Y\cap Cl_X(S).$ Then $p\not \in Cl_X(S)$ because $$\{p\}\cap Cl_X(S)= (\{p\}\cap Y)\cap Cl_X(S)=\{p\}\cap (Y\cap Cl_X(S))=\{p\}\cap S=\emptyset.$$ So let $f:X\to [0,1]$ be continuous with $f(p)=0$ and $f(Cl_X(S))\subset \{1\}.$ Then $f|_Y:Y\to [0,1]$ is continuous, with $f|_Y(p)=f(p)=0$ and $f|_Y(S)=f(S)\subset f(Cl_X(S))\subset \{1\}.$

(2). A subspace of a normal space can fail to be normal.

Example: Let $Y$ be a locally compact, non-normal Tychonoff space. Let $T_Y$ be the topology on $Y.$ Note that $Y$ is not compact because compact Hausdorff spaces are normal. Take a point $p\not \in Y$ and let $X=\{p\}\cup Y.$ Let the topology $T_X$ on $X$ be generated by the base $$T_Y\cup \{\{p\}\cup S: S\subset Y\land X\backslash S \text { is compact}\}.$$ The set $X$ with topology $T_X$ is called the Alexandroff (or one-point) compactification of $Y$, denoted $aY$ or $\alpha Y.$ You may verify that

(i). $X$ is a compact Hausdorff space and therefore a normal space.

(ii). $T_Y=\{t\cap Y: t\in T_X\},$ so $T_Y$ is the subspace topology on $Y,$ as a subspace of $X.$

So $Y$ is a non-normal subspace of the normal space $X.$

(3). Here is one you may have overlooked: A space $X$ is an Uryssohn space iff when $p,q$ are distinct points of $X,$ there are open sets $U_p,U_q$ with $p\in U_p$ and $q\in U_q,$ and $\overline {U_p}\cap \overline {U_q}=\emptyset.$ A space can be Hausdorff and not Urysohhn. ( We could call an Uryssohn space a $T_{2\frac {1}{2}}$ space but nobody does.) Q: Is every subspace of an Urysohhn space also Uryssohn?

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    You can also consider the classical example of the Tychonoff plank for non-normality of subspaces. We can use $\Psi$-space for your locally compact non-normal space, e.g. – 2017-01-06
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    @HennoBrandsma. Yes of course. In an answer to a basic Q like this one, I usually try to keep it elementary. – 2017-01-06
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    $T_{2\frac{1}{2}}$-space is already used by some for a functionally Hausdorff space: for every pair of distinct points $x \neq y$ there is a real-valued continuous function $f$ on $X$ with $f(x) = 0$ and $f(y) =1$, in analogy with Tychonoff being $T_{3\frac{1}{2}}$, I suppose. But a $T_3$ space need not be functionally Hausdorff, so I usually avoid the $T_{2\frac{1}{2}}$-term, as I prefer $T_j$ to imply $T_i$ when $i < j$.. – 2017-01-06