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It seems like it would be very simple, but I’m having trouble proving the following:

Let $X =\{x\in \mathbb{R}^n\,: ||x||_1\}$ and $Y=\{y\in \mathbb{R}^n\,: ||y||_2=1\}$. Prove that $(X,||.||_2)$ and $(Y,||.||_2)$ are homeomorphic to each other.

I’m trying to find a homeomorphism in spaces using the same metric $||.||_2$ because I was trying to demonstrate that $X$ was compact in Euclidean space. (I know there are easier ways to answer this but it just came up when I was trying to prove it by showing that: “X is totally bounded and complete” or that “every sequence in X has a convergent subsequence”).

Basically, I’m having trouble finding a function from one to the other (I think I can show that it’s a homeomorphism after that though). Could anyone come up with one?

More generally, do you have any suggestions on finding functions between metric spaces before checking if they are homeomorphisms? Once I am given a function mapping one metric space to another, I can generally check whether it fulfills the necessary conditions to be a homeomorphism, but I often seem to have trouble finding the function itself.

Related to the question above, how would one prove that $(\mathbb{R}^n, ||.||_1)$ and $(\mathbb{R}^n, ||.||_2)$ are homeomorphic?

Edit: I originally proved it (I think) using sequential compactness (See below), but the question of whether there was a homeomorphism from X to Y came up while I was thinking of alternative ways to do so.

Carothers 8.8: Prove that the set $\{x\in \mathbb{R}^n\,:\, ||x||_1 = 1\}$ is compact in $\mathbb{R}^n$ under the Euclidean norm.

akm
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    Would you kindly review your notation? Your definition of $Y$ is a little off, how is $x_2$ related to $y$?. I'm sure improving this aspects will help you to receive more attention. For the last question, if $||.||_1$ and $||.||_2$ are arbitrary metrics, the result may not hold. – AHandsomeAlien May 09 '20 at 07:27
  • Sorry about that! Careless error. Hope it makes more sense now. – akm May 09 '20 at 15:12
  • By $||.||_1$ and $||.||_2$ I meant the “taxicab” and “Euclidean” norms – akm May 09 '20 at 15:13
  • You can show that $X$ is compact by noticing it is the pre image $f^{-1}(1)$ of a closed set by a continuous functions $f=||.||_1$ and the same goes for $Y$. For the last question, one generally prove that they are homeomorphic by showing they have the same topology, which is the one induced by the norms, in other words, if the norms are equivalent they are homeomorphic. – AHandsomeAlien May 09 '20 at 15:26
  • Could I assume that there existed functions from X to Y? Intuitively I could imagine one (eg mapping points in the perimeter of a diamond onto the perimeter of a circle) but wasn’t sure if I had to come up with one before making that assumption formally. – akm May 09 '20 at 15:30
  • I see what you’re saying re compactness. But I was just curious, separately, about how to show that X and Y are homeomorphic. – akm May 09 '20 at 15:35
  • Is your problem stated correctly? $X$ is the unit sphere with respect to norm 1 and $Y$ is the unit sphere for norm 2, however you want to show that $X$ with the topology induced by the norm 2 is homeomorphic to $Y$ with topology induced by the norm 2? – AHandsomeAlien May 09 '20 at 15:42
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    [check this answer] (https://math.stackexchange.com/questions/3215832/can-one-argue-homeomorphism-via-equivalence-of-metrics). – AHandsomeAlien May 09 '20 at 15:49
  • Correct. I was seeing X as the set of points that would meet the criteria for $||x||_1 =2$ but in a metric space using the Euclidean metric. At least that was my intuition related to the original problem, which only gave me the set X without a metric, but then asked to prove it as compact in Euclidean space. (Original problem now linked in my first post). – akm May 09 '20 at 15:52
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    Thanks for that answer. I can see how they are equivalent, and there are a bunch of problems in textbooks saying that it’s trivial to prove the of real numbers with the metrics $||x||1$, $||x||_2|$, $||x||{\infty}$ Are homeomorphic but I always had a hard time proving this. I’ll look at the answer more carefully to answer my second question. Thanks! I’m not sure if my first question is still clear, I can try to clarify further. – akm May 09 '20 at 15:57
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    In the comment two above, I meant $||x||_1=1$ not $2$ – akm May 09 '20 at 15:58

2 Answers2

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Let $f: X \rightarrow Y$ by $$f(x_1, \cdots, x_n) = \left(\frac{x_1}{\sqrt{x_1^2 + \cdots + x_n^2}}, \cdots, \frac{x_n}{\sqrt{x_1^2 + \cdots + x_n^2}}\right)$$ which is a homeomorphism.

cha21
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The only function $X \to Y$ I can think of is $(x_1, x_2, ..., x_n) \mapsto (x_2, x_1, ..., x_n)$. If you can only think of one, it is the correct one most of the time. It is clear that it is self-inverse, so you only need to show it is continuous. Hint: projections and inclusions are continuous. Actually $X$ is not compact, consider the cover $(\{1\} \times ]-n, n[ \times ... \times ]-n, n[)_{n \in \mathbb{N}}$.

Edit: didn't see your second question. You would prove that a basis element in $(\mathbb{R}^n, ||.||_1)$ is open in $(\mathbb{R}^n, ||.||_2)$ and vice versa. You would need a description of the metrics to do this though.

user388557
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  • Sorry I think I used poor notation. I was trying to demonstrate that the set $X$, where the sum of the absolute value of each point’s coordinates equals 1 $ \sum_{i =1}^n |x^i| =1$— basically a diamond around zero — was compact in Euclidean space. I wanted to avoid using the “closed and bounded” definition of compactness and thought of this question when trying to prove that $(X, ||.||_2)$ ie X with the Euclidean norm as a metric, was complete and totally bounded. – akm May 09 '20 at 15:26