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Is there any topological Property $P$ if it is satisfied on a subspace topology $N$ of a topological space $X$, then it must be satisfied on a topological space $X$. Thanks.

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Please, try to make the title of your question more informative. E.g., Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here. –  Martin Sleziak May 18 at 8:26
    
Is there a motivation behind this question? –  Najib Idrissi May 18 at 9:47

3 Answers 3

Not first-countable. Not second-countable. Not Hausdorff. Not regular. Not completely regular. Not T0. Not T1. Not metrizable.

In general, if $\mathfrak{P}$ is a topological property inherited by subspaces, then the topological property not-$\mathfrak{P}$ is of the kind sought.

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Recall that a topological property is called hereditary if the following holds: If $X$ has the property P and $N$ is subspace of $X$, then $N$ also has the property P.

If you take any hereditary property P, the negation of it $\neg P$ will fulfill the conditions you ask in your post.

In fact, the property $P$ is hereditary if and only if $\neg P$ has fulfills the requirements from your post.


In some contexts, results of this type are might be called reflection theorems. However, as the two formulations are equivalent, they can be formulated as in your post, or dually. Quite often some additional condition is given on the subspace.

The following quote is from Encyclopedia of General Topology; Edited by: Klaas Pieter Hart, Jun-iti Nagata and Jerry E. Vaughan, Elsevier, 2003, p.16.

Generally speaking, a reflection theorem is a theorem of the form "if a set (or a space) $X$ has a property $P$, then there is a subset (or a subspace) $Y$ of a small size (in some sense) satisfying $P$.

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Do you consider having a nontrivial open set to be a topological property? $N \not = \emptyset$ implies $X \not = \emptyset$, so $N$ having a nontrivial open set implies that $X$ does too. You could cook up other "properties" like this that satisfy your answer, but in a boring way. (Like that $X$ contains a property $P$ subspace.)

But in general I don't think so - if I can write $X$ as the disjoint union of $N$ and $M$, then most properties I can think of have to be satisfied by $N$ and $M$ in order to be satisfied by $X$ (the separation axioms, for instance). For others (such as connectedness), this is not enough.

But the upshot is that if you hand me a property that a space $N$ has and a space $M$ doesn't, then I think $N \dot \cup M$ is unlikely to have that property. This is especially so if that property can be characterized locally, or by some global characteristic - like that a space being compact implies that every closed set is compact.

I would like to be proved wrong though.

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Thanks very much to all. –  user132983 May 18 at 9:33

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