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One example was given in Munkres. Take the order topology on $\mathbb{R}^2$ using the dictionary order. Then the subspace topology on the unit square is not the same as the "ordered square" on the same subset.

Are there any criterion for when the subspace topology always equals the topology defined on the subset, or when they are not equal? Also, what are some other examples where they are not equal?

user95041
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3 Answers3

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Yes. There is a theorem in Munkres that states that for a convex subset of an ordered set, the order topology on the subset coincides with the subspace topology. (Convex means that if $a$, $b$ are in the subset, so is $[a,b]$.) The ordered square is not a convex subset of $\mathbb R^2$, so the two topologies don't have to coincide. A simpler example is the set $A=[0,1)\cup\{2\}\subset \mathbb R$. The order topology gives that $A$ is homeomorphic to $[0,1]$, whereas in the subspace topology $\{2\}$ is an open set.

Cheerful Parsnip
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  • Excellent, thank you. Are there similar criteria with respect to metric topology? – user95041 Oct 08 '13 at 21:57
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    If $X$ is a metric space and $A\subset X$ has the induced metric, then the subspace and metric topologies coincide. – Cheerful Parsnip Oct 08 '13 at 22:05
  • They always coincide? Good, this explains why this subtlety has been ignored (because it doesn't matter) in my analysis classes. – user95041 Oct 08 '13 at 22:10
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    @user95041: There are many subtleties which often are not mentioned because it is almost trivial. For example, if $A\subseteq X$ and $B\subset Y$, then the product $A\times B$ (with its product topology) is a subspace of $X\times Y,$ quite as you would expect it. – Stefan Hamcke Oct 08 '13 at 22:32
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Let $\langle X,\tau\rangle$ be any topological space, and let $Y$ be any subset of $X$ with at least two points. Let $\tau_Y$ be the subspace topology on $Y$. Then we can always find a topology $\tau'$ on $Y$ such that $\tau'\ne\tau_Y$. A very simple way to do it is to notice that if $|Y|\ge 2$, the discrete and indiscrete topologies on $Y$ are distinct, so at least one of them must be different from $\tau_Y$.

If we already have some topology $\tau'$ on $Y$, the only time that we can be certain (without actually knowing what the topologies are) that $\tau'=\tau_Y$ is when $|Y|=1$.

Brian M. Scott
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  • Sorry in advance if I'm not reading your comment correctly, but I'm asking specifically about defining the "same" topology on the superset and have it be inherited by the subset as a subspace topology, versus defining that "same" topology on the subset directly. So in your example, taking discrete and indiscrete topologies would not be defining the "same" topology on that subset. – user95041 Oct 08 '13 at 21:28
  • In the Munkres example, the dictionary order topology was inherited by the unit square, and that was shown to be unequal to the order topology defined directly on the unit square (called the "ordered square"). – user95041 Oct 08 '13 at 21:29
  • @user95041: In that case your question probably isn’t answerable: you’d have to define exactly what same means here, and I’m not at all sure that this is possible. – Brian M. Scott Oct 08 '13 at 21:33
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I suppose you are thinking of something along the lines of the following:

Take a space $(X,\tau)$ with an equivalence relation $\sim$. If $(A,\tau_A)$ is a subspace of $X$, then we can restrict the relation $\sim$ to the relation $\sim_A$ on $A$. On the quotient set $X/\sim$, the set of all equivalence classes, the topology consists of all sets $U$ such that $q^{-1}(U)$ is open in $X$, where $q:X\to X/\sim$ is the quotient map.
Now, the most natural guess for the topology on $A/\sim_A$, the set of all $\sim_A$-equivalence classes in $A$ (which is in a natural way equivalent to the set $q(A)$, the set of all $\sim$-equivalence classes in $X$ with at least on point from $A$), would be the quotient topology again. But: this quotient topology on $A/\sim_A$ is not always the same as the topology inherited from $X/\sim$ (it is always finer, though).

To put it briefly

the quotient topology on the subspace is finer than the subspace topology of the quotient.

Or we can formulate it

The restriction of a quotient map need not be a quotient map .

Here is an example: Let $X=[0,1]\times[0,1]$ with the usual topology. Let $Y=X/B$, the space obtained by collapsing the lower edge $B:=[0,1]\times\{0\}$ to a point, and $q:X\to Y$ the quotient map. Let $A=[0,1]\times(0,1]\ \cup\ \{(1,0)\}$. Now the set $U=(1/2,1]\times I\cap A$ is open in $q(A)$, but no open set in $X/\sim$ can intersect $q(A)$ in $U$ because its preimage had to be an open set containing $I\times\{0\}$.

See also my answer in Finding a counterexample; quotient maps and subspaces for another counterexample, as well as Ronnie Brown's answer for a criterion for equality of the topologies.

Community
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Stefan Hamcke
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  • I'm a little fuzzy with the quotient topology but if I'm reading you right, yes, this is in the spirit of what I'm trying to get at. I was hoping for more simple examples involving the order topology, for example, but thanks for your contribution. – user95041 Oct 08 '13 at 21:52
  • I thought you expected something not only involving the order topology, but an example where the topology on the subspace is defined in the same way but is not the subspace topology. That's how I understood your question. Never mind. I think the interesting point is that the subspace topology can be finer (as in your example), but it can also be coarser (as in my case). Talking of examples, I will provide a concrete example. – Stefan Hamcke Oct 08 '13 at 21:58
  • No, you are quite right. It is exactly what I was asking for, only I was "hoping" for more simple examples involving more familiar topologies due to my limited understanding of the quotient topology. – user95041 Oct 08 '13 at 22:04
  • @user95041: Ah, okay, :-) – Stefan Hamcke Oct 08 '13 at 22:15