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In a text I am reading, the author defines two norms on a vector space $X$ to be equivalent if they induce the same topology on $X$. The text does not define what it means however for two topologies to be equivalent. The only definition I can think of that seems reasonable is that two topologies $T_1$ and $T_2$ are equivalent if an open set in $T_1$ is an open set in $T_2$ and conversely. Since open sets in a normed space are determined by the norm-relative open balls, this would mean that any open ball about a point $x$ with respect to one norm not only contains an open ball about $x$ with respect to the other norm but is also contained within an open ball relative to that other norm. Is this the right way to look at the situation? Is my definition of "equivalent toplogies" appropriate in this context?

ItsNotObvious
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    Your proposed definition of equivalence for topologies coincides with equality. – Mariano Suárez-Álvarez Nov 04 '11 at 14:10
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    To follow up on Mariano Suárez-Alvarez's comment, in contexts like this, equivalence is a relation on metrics (or on norms), not on topologies. Also, there are several types of such equivalences in texts and in the literature, with names that are sometimes not very informative or in agreement with other names in the literature. I posted an essay about three such notions, which I called "Lipschitz equivalent", "uniformly equivalent", and "topologically equivalent" about 5 years ago at http://groups.google.com/group/sci.math/msg/9e825cd2be094cd7 – Dave L. Renfro Nov 04 '11 at 14:27
  • @MarianoSuárez-Alvarez So, basically my definition of "equality" is equivalent to, pun intended, to equality of sets. Which, if I had read my own definition a little more carefully I would have understood! – ItsNotObvious Nov 04 '11 at 15:27
  • @DaveL.Renfro Thanks for the reference; that's helpful – ItsNotObvious Nov 04 '11 at 15:28
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    Indeed, in your first sentence you didn't writ "equivalent topologies", you wrote "the same topology". This is literally what is meant. – Nate Eldredge Nov 04 '11 at 18:25
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    @DaveL.Renfro: I ran into the same issue in my question http://math.stackexchange.com/questions/5315/notions-of-equivalent-metrics – Nate Eldredge Nov 04 '11 at 18:26
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    @3Sphere: There is a similar discussion here: http://at.yorku.ca/cgi-bin/bbqa?forum=ask_a_topologist_2008;task=show_msg;msg=2219.0001 – Karatuğ Ozan Bircan Nov 04 '11 at 19:03
  • It's worth noting that since here we're working with normed vector spaces, if they're homeomorphic they'll automatically be uniformly homeomorphic; we don't need to distinguish those notions here. Of course we certainly do for more general metric spaces or uniform spaces... – Harry Altman Nov 06 '11 at 09:04

3 Answers3

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Given two topologies $\tau,\tau'$ on a set $X$, we say $\tau$ is coarser (or weaker) than $\tau'$ iff $\tau\subset\tau'$. Accordingly we say $\tau'$ is finer (or stronger) than $\tau$.

Equivalently, this means $\tau$ has more (read:not less) opens, or equivalently $\tau$ has more (read:not less) neighborhoods, or equivalently the identity set map $(X,\tau)\to (X,\tau')$ is continuous, or equivalently for each subset the closure in $\tau$ is included in the closure in $\tau'$.

Thus the following are equivalent ways of saying that $\tau,\tau'$ are the same:

  • $\tau=\tau'$

  • each of $\tau$ and $\tau'$ is coarser/finer than the other

  • $\tau,\tau'$ have exactly the same opens

  • $\tau,\tau'$ have exactly the same neighborhoods

  • for each subset, its closures in $\tau$ and in $\tau'$ coincide.

  • the identity set map $(X,\tau)\to (X,\tau')$ is a homeomorphism.

wildildildlife
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  • Do you happen to have any references for the second paragraph? I'm self-studying intro level topology and I want to see how those come about. – S.D. Aug 11 '13 at 11:35
  • @ShantDanielian: all these statements follow more or less directly from the definitions, so you could regard them as exercises. – wildildildlife Aug 12 '13 at 15:16
  • Ah, I see. Thanks, I'll try to work them out. – S.D. Aug 12 '13 at 22:39
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    I think if $\tau\subset\tau'$, then $\tau'$ has more opens. But in your second paragraph, you said $\tau$ has more opens. Also, if $\tau\subset\tau'$, then I think the identity map $(X,\tau')\to (X,\tau)$ is continuous. – user398843 Feb 21 '18 at 04:09
  • @wildildildlife: I think your second paragraph there is wrong (as mentioned by @user398843). Cheers! – asn32 Mar 18 '18 at 18:30
  • @asn32 It seems that the author doesn't use this account anymore. I just edited the answer. Changes will be posted after peer review. – user398843 Mar 18 '18 at 19:37
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Two norms are equivalent if they induce the same topology. So it does not matter whether the author of your book defines the notion of "equivalent topologies", since he does not need it (at least not here). I.e., you work here with equality of topologies. (Which is reflexive, symmetric and transitive, so you may call the same topologies equivalent, if you wish. However, the name "equivalent" would be redundant.)

You also asked whether this:

open set in $T_1$ is an open set in $T_2$ and conversely

is the same as equality of topologies. Indeed, it is. Every set open in $T_1$ being open in $T_2$ is inclusion $T_1\subseteq T_2$. The converse implication gives you the converse inclusion.

BTW I guess all of this was already said in comments.

Martin Sleziak
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Two topologies are called equivalent if any non-empty open set of the first topology contains a non-empty open set of the second topology and, conversely, every non-empty open set of the second topology contains non-empty open sets of the first topology.

This does NOT mean that the two topologies are the same. So give me my two vote down back! :)

In case of normed spaces, this condition can be written in a quite nice way: two norms $||\cdot||_1$ and $||\cdot||_2$ on $X$ are equivalent if and only if there are two positive numbers $C_1,C_2>0$ such that, for all $x\in X$,

$$ C_1||x||_1\leq||x||_2\leq C_2||x||_1 $$

you can verify by yourself that this condition is equivalent at the one between topological spaces just remembering what is a base of open sets of the topology induced by a norm.

Valerio Capraro
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    Re: first sentence. I never heard this notion of "equivalent" used for topologies (see also Dave's comment). Why introduce a new word for something that's already aptly described by "equality", as Mariano pointed out? – t.b. Nov 04 '11 at 14:52
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    I have never seen this definition of the equivalence of two topologies. In the case of topologies coming from norms on a vector space, two topologies are equivalent according to your definition if they are the same. For general topologies I find this definition rather odd. Do you have any source for this definition? – Stefan Geschke Nov 04 '11 at 19:45
  • @StefanGeschke: This "definition" of equivalence gives you that two topologies are equivalent when they are equal. That is what t.b. is pointing out. – André Caldas Nov 05 '11 at 00:33
  • @Andre Caldas: Are you sure about this? In the general case, I mean. For topological vector spaces I agree. But for general topologies the natural notion of equivalence (which then turns out to be equivalent to equality) would be this: Two topologies are equivalent if for each point $x$ and open neighborhood $U$ of $x$ in the first topology there is an open set $O$ in the second topology such that $x\in O\subseteq U$ and vice versa. – Stefan Geschke Nov 07 '11 at 08:31
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    I had to think about this for some time, but here is an example: Consider two topologies on the real line. Let $\tau_0$ be the usual topology, generated by the open intervals. Let $\tau_1$ be the topology generated by the intervals of the form $(a,b]$, $a<b$. Every nonempty set in $\tau_0$ contains a nonempty set in $\tau_1$ and vice versa, yet the two topologies are not the same. There is also the trivial problem that any two topologies on a space are equivalent in Valerio's sense since they both contain the empty set. But my example shows a deeper "problem" with the definition. – Stefan Geschke Nov 07 '11 at 08:49
  • @StefanGeschke: You are right. I did not pay much attention to Valerio's definition. Since he was talking about topological vector spaces, I missread his statement like this: "every open neighborhood of $0$ form topology 1 contains an open neighborhood of $0$ from topology 2 and vice-versa". This would imply that the topologies are the same. You are right, I had not noticed that he does not talk about the $0$ or neighborhoods of any given point. – André Caldas Nov 07 '11 at 16:57
  • So, the end of the story is that I gave a right definition, but it has two vote down... :( – Valerio Capraro Nov 07 '11 at 17:29
  • @Valerio: Can you give a reference for this strange definition of equivalence? I've never seen it before. – Cheerful Parsnip Nov 07 '11 at 18:06
  • boh.. just remember from undergraduate school. – Valerio Capraro Nov 07 '11 at 18:36
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    @Valerio Are you sure it wasn't equivalence of bases? See e.g. this book, this book this thread – Martin Sleziak Nov 07 '11 at 21:24
  • what you say means that $\tau=\tau'$. $;$ since if $\frak{B}$ is a basis for $(X,\tau)$ and $\frak{B}'$ is a basis for $(X,\tau')$. $;$then two condition $(1),(2);$are equivalent. $(1):;$$\tau\subseteq \tau';$ $(2):;\forall x\in X;$ and every $B\in\frak{B}$ containing $x$, there is a $B'\in\frak{B}'$ such that $x\in B'\subseteq B$.$;$ now if we consider $\frak{B}=\tau$ and $\frak{B}'=\tau'$ then the conclusion is that two condition $(1'),(2')$ are equivalent. $(1');$$\tau\subseteq \tau';$ $(2')$ for every $;U\in\tau;$ there is a $;U'\in\tau'$ such that $;U'\subseteq U$. – Amirhossein Haddadian Feb 19 '23 at 22:53