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My lecture notes read "Let $(X,d_1)$ and $(X,d_2)$ be metric spaces with the same underlying set $X.$ Then $d_1$ and $d_2$ are called topologically equivalent if the identity map is continuous as a map from $(X,d_1)$ $\rightarrow (X,d_2)$ and as a map from $(X,d_2) \rightarrow (X,d_1).$"

I'm new to Topology and Metric Spaces and this doesn't make much sense to me. All we've really done so far in this module is introduce metric and normed spaces, generalised things like open and closed sets, uniform continuity, some new concepts like interior, neighbourhood, closure etc. but this definition makes no sense to me. I understand subspaces and why you can have a set and a metric that make up a metric space, and say you take a subset of that set you can have a subspace by restricting the metric to that subset. However, at the points in the subset the restricted metric is equal to the larger metric. But, this definition really does make no sense to me. Idk how to even visualise it because I visualised the metric as a mapping. How would you even demonstrate topological equivalence using this definition? It just really makes no sense to me.

KeynesianSpaceman
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1 Answers1

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For a set $X$ with two topologies $\tau$ and $\tau'$ (collections of subsets called open sets such that some properties hold -- do you know what a topology is on a space, or only what metrics and norms are?), saying the identity map from $(X,\tau)$ to $(X,\tau')$ is continuous is just saying every open set for $\tau'$ is open for $\tau$ (yes, in that order), so saying the identity maps $(X,\tau) \to (X,\tau')$ and $(X,\tau') \to (X,\tau)$ are both continuous is just saying the two topologies $\tau$ and $\tau'$ are the same: a subset of $X$ is open in one topology exactly when it is open for the other. Passing to complements, this property is equivalent to saying the two topologies define the same closed subsets in $X$.

When you're dealing with a metric space $(X,d)$, a closed subset is a subset $C$ that contains the limit of every convergent sequence in $C$, so saying two metrics have the same closed subsets is the same as saying the two metrics define the same notion of convergence in $X$.

Example. Let $X = C[0,1]$ be the set of all continuous functions $[0,1] \to \mathbf R$. On this space here are two norms: $||f||_\infty = \max_{0 \leq x \leq 1} |f(x)|$ and $||f||_1 = \int_0^1 |f(x)|\,dx$. These are not topologically equivalent: if $||f - f_n||_\infty \to 0$ then $||f-f_n||_1 \to 0$ because $||f-f_n||_1 \leq ||f-f_n||_\infty$, but there are convergent sequences for the norm $||\cdot||_1$ that do not converge for the norm $||\cdot||_\infty$. For example, let $f_n(x) = x^n$ on $[0,1]$. Then $||f_n||_1 = \int_0^1 x^n\,dx = 1/(n+1) \to 0$ as $n \to \infty$, but $||f_n||_\infty = 1$ for all $n$, so $f_n \to 0$ in the metric defined by $||\cdot||_1$ but $f_n \not\to 0$ in the metric defined by $||\cdot||_\infty$. It turns out that $f_n$ does not converge at all in the metric defined by $||\cdot||_\infty$, but proving that requires more work (just because it doesn't have limit $0$, why couldn't it have some other limit in $C[0,1]$?). See https://kconrad.math.uconn.edu/blurbs/analysis/metricspaces.pdf, especially Examples 2.3 and 3.11

KCd
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  • Thank you, I don't really understand because I haven't heard the definition of convergence you gave before. The difficulty I'm running into really is the distinguishment between the metrics in this case. I can understand the open ball definition because that's relational to epsilon-delta definitions, but the fact the metrics are different is what is confusing me; and the fact that this isn't even the map in this regard. Maybe I'm thinking about it wrong since I'm thinking of a map as a function like x^2, or x^3 etc. and the notion of different metrics is confusing me – KeynesianSpaceman Jan 16 '23 at 20:32
  • Also, we haven't done topology yet, only metrics and norms – KeynesianSpaceman Jan 16 '23 at 20:33
  • For example, the example we are given of topological equivalence is d_1(x,y) = |x - y| and d_2(x,y) = |e^x - e^y| and the set X is R. The issue I'm having here is really trying to understand what the hell the metric and the map are even saying. So in our case I'm assuming that we are trying to say that for d_1(x,y) < delta that implies d_2(x,y) is less than epsilon. But I don't even know what the identity map has to do with this at all. – KeynesianSpaceman Jan 16 '23 at 20:40
  • For $X$ with metrics $d$ and $d'$ on it, that $f \colon (X,d) \to (X,d')$ is continuous at $a \in X$ means for all $\varepsilon > 0$ there is $\delta > 0$ (possibly depending on $\varepsilon$ and $a$) such that if $d(a,x) < \delta$ then $d'(f(a),f(x))< \varepsilon$: each $d'$-open ball around $f(a)$ contains the image of a $d$-open ball around $a$. So if $f$ is the identity mapping, its continuity means: for all $a \in X$ and $\varepsilon > 0$ there is $\delta > 0$ such that if $d(a,x) < \delta$ then $d'(a,x)< \varepsilon$: each $d'$-open ball around $a$ contains a $d$-open ball around $a$. – KCd Jan 16 '23 at 20:47
  • I understand what you mean in that sense in terms of the generalisation, I just have no idea about the application of this to actual examples, I think this is what I'm lacking perhaps – KeynesianSpaceman Jan 16 '23 at 20:55
  • Consider $d_1((x,y),(x',y')) = \sqrt{(x-x')^2+(y-y')^2}$ and $d_2((x,y),(x',y')) = \max(|x-x'|,|y-y'|)$ on $\mathbf R^2$. These are both metrics. An open ball in the 1st metric is the usual open ball in the plane, but an open ball in the 2nd metric is an open square (the inside, not including its boundary) with sides parallel to the axes. That these two metrics are equivalent is "obvious": a circular region around a point contains a smaller square region around that point (with sides parallel to the axes) and a square region around a point contains a smaller circular region around that point. – KCd Jan 16 '23 at 20:59
  • This is what I don't understand in the sense that in what way is this an identity map going on here? Or how could you apply that to here? What exactly is getting mapped to itself? Because it isn't (x,y) because d_1 doesn't give x back. We talk a lot about f(a) etc. in previous comments. What exactly is the f here I suppose is what I'm asking – KeynesianSpaceman Jan 16 '23 at 21:20
  • The function $f \colon \mathbf R^2 \to \mathbf R^2$ where $f(x,y) = (x^2y, 7xy - 8y^3)$ is continuous, right? The identity function on a set is just the function that fixes everything: $f(x,y) = (x,y)$ for all $(x,y)$ in $\mathbf R^2$. – KCd Jan 16 '23 at 23:57