If $X$ is a topological space then it's (covering) dimension is defined as a minimal number $n$ such that for every finite open cover $\{U\}$ of $X$ there is a finite open cover $\{V\}$ of $X$ that refines $\{U\}$ and such that every point $x \in X$ is contained in no more than $n+1$ set of $\{V\}$.

If $X$ has a dimension $n$ and $F$ is a closed subspace of $X$ then $\dim F \leqslant n$. It is an easy exercise: if $\{U\}$ is an open cover of $F$ then $\{U\}$ and $F^c$ give and open cover of $X$ and we can use the definition of dimension for $X$ to obtain $\dim F \leqslant \dim X$.

But I didn't find in any book that it is true in general. Also I didn't find a counterexample. Is it possible that a subspace $Y$ of $X$ has a dimension greater then dimension of $X$? If it is possible it is very interesting to look at such example.


There are examples of T$_{3 \, 1/2}$-spaces $X$ and closed $F \subseteq X$ with $\mathrm{dim}(F) > \mathrm{dim}(X)$.

Problem 7.4.6 (p.419) of Engelking's General Topology (1989 ed.) asks for an example of such a space. This example is therein credited to

Yu. M. Smirnov, On the dimension of proximity spaces (Russian), Mat. Sb. N.S. 38 (1956), 283–302 MR0082095 (English translation Amer. Math. Soc. Transl. (2) 21, 1962, 1–20. MR150728).

For non-closed subsets, it is a bit easier (though I will again steal from Engelking).

Take a zero-dimensional not strongly zero-dimensional space $X$ (e.g., Dowker's Example, 6.2.20 in Engelking). This space has a zero-dimensional compactification $\gamma X$ (Corollary 6.2.17) and every compact — in fact Lindelöf — zero-dimensional space is strongly zero-dimensional (Theorem 6.2.7). Therefore $ \mathrm{dim} ( \gamma X ) = 0 < \mathrm{dim} (X)$.

Added: I see now that the definition of the covering dimension given applies only to normal spaces. As stated in the OP, it is a theorem that closed subspaces of normal spaces cannot have larger covering dimension.

For the second example the space $X$ from Dowker's Example is normal, and clearly so, too, is its zero-dimensional compactification $\gamma X$, and so this would yield an example of a normal subspace of a normal space with strictly larger covering dimension.

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    $\begingroup$ If we use the definition given in my post (it is from Alexandrov,Pasynkov "Introduction to dimension theory", Александров, Пасынков "Введение в теорию размерности") then the dimension of closed subspace is always not greater then dimension of the whole space (it is a theorem from the above book (chapter 2,paragraph 2, theorem 1 "about monotonicity of dimension with respect to closed subspaces") $\endgroup$ – Appliqué Mar 26 '13 at 13:16
  • $\begingroup$ @Nimza: I am certainly no expert in dimension theory — one of a few reasons my answer was community wiki — but I think I see my error. The definition you gave works for normal spaces but there is a generalisation that can be applied to completely regular spaces (and yields the same result for normal spaces). Yes, a closed subspace of a normal space can never have larger covering dimension. (But the second example would still suffice to show that arbitrary normal subspaces of normal spaces can have larger covering dimension.) $\endgroup$ – user642796 Mar 26 '13 at 13:31
  • $\begingroup$ The definition given in my post is introduced for general topological spaces. Some theorems are proved in general setting in the Alexandrov's book. After author says that "now we will restrict our considerations to normal spaces", but not before this theorem. (I think that there may be two generalisations of the above definition of covering dimension for normal spaces to more general classes of spaces, one is via my definition applied to general spaces) $\endgroup$ – Appliqué Mar 26 '13 at 13:46
  • $\begingroup$ @Nimza: Please feel free to edit (fix) this answer. You clearly know much more about this topic than I do. (Though I must say that I have learned some nice things today. Thanks for that!) $\endgroup$ – user642796 Mar 26 '13 at 13:57

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