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6 votes

A complete metric space contains a convergent sequence or an infinite discrete subset

I. With Ramsey's theorem. Ramsey's theorem says that, for any positive integer $r$, if the $r$-element subsets of an infinite set are colored with two colors, then there is an infinite subset whose $r$...
user14111's user avatar
  • 1,543
4 votes

A complete metric space contains a convergent sequence or an infinite discrete subset

The claim in the question can be shown using  the infinite version of Ramsey theorem. I.e., we are using the following fact: If $A$ is an infinite sets and all 2-element subsets of $A$ are colored ...
Martin Sleziak's user avatar
3 votes

Can the twice-punctured plane be given a homogeneous metric?

Not a complete answer. Let me restrict attention to Riemannian metrics; I don't know what the non-Riemannian case looks like. If a Riemannian surface is homogeneous then in particular it must have ...
Qiaochu Yuan's user avatar
2 votes
Accepted

If $f:X \rightarrow Y$ is an $\epsilon$-map between compact metric spaces then there exists a $\delta > 0$ such that $diameter(f^{-1}(Z)) < \epsilon$

Suppose for the sake of contradiction that no such $\delta$ existed. Then, for all natural numbers $N$, there exist sets $Z_n$ such that $\operatorname{diameter} Z_n < \frac{1}{n}$, but $\...
Theo Bendit's user avatar
  • 51.6k
2 votes
Accepted

Confusion about the criteria for a complete metric space to be a length space

They have a webpage with errata for the book here. In the errata they say the following regarding sections 2.4-2.5: "Although we usually allow infinite distances, many statements here are ...
Moishe Kohan's user avatar
1 vote

Prove that there are no onto functions like $f:[0,1]\rightarrow[0,1]$ such that for each $x\in[0,1]$, $f^{-1}(x)$ is open.

Assume that $f:[0,1]\rightarrow [0,1]$ is surjective and, for each $x\in[0,1]$ $f^{-1}(x)$ is open. As mentioned in the comments, each open neighborhood in $[0,1]$ contains at least one rational ...
user140776's user avatar
  • 1,953
1 vote

Parametric Equation of a unit circle when the angle between x-axis and y-axis is not 90 degrees

Let's consider a non orthogonal Cartesian reference system with unit vectors of the axes $\textbf{e}_1,\textbf{e}_2$, with $|\textbf{e}_1|=|\textbf{e}_1|=1$ and $\textbf{e}_1\cdot\textbf{e}_2=\cos\...
Vincenzo Tibullo's user avatar
1 vote

Prove that there are no onto functions like $f:[0,1]\rightarrow[0,1]$ such that for each $x\in[0,1]$, $f^{-1}(x)$ is open.

Do we really need the assumption that $f$ is surjective to derive contradiction? It seems to me that any function $f\colon [0,1]\to[0,1]$ where $f^{-1}(x)$ is open for all $x\in [0,1]$ must be ...
dusko's user avatar
  • 11
1 vote

Are there compact metric spaces with Hausdorff measure equal to 0 or infinty

It might be possible with $\{0,1\}^{\mathbb{N}}$, if you try different metrics on it. Given $a\in (0,\infty)^{\mathbb{N}}$ and $\gamma>0$ put $d_{a,\gamma(x,y)}=a_{\min\{n\in\mathbb{N}\mid x(n)\neq ...
dialegou's user avatar
  • 237
1 vote

Give example of $f$ that is open but neither closed not continuous (in 2D).

I am reading "Introduction to Topological Manifolds Second Edition" by John M. Lee. This problem is the same problem as Problem 2.5(a) in this book. I solved this problem as follows: Let $X=...
佐武五郎's user avatar
1 vote

Understanding the Proof of Relative Openness Theorem in Rudin's Principles of Mathematical Analysis

This may be an approach to consider (expanding somewhat on the comment by @MPW). Focusing on the "right-to-left" part of the proof: Suppose $G$ is open in $X.$ This means that every point ...
ad2004's user avatar
  • 1,866
1 vote

Understanding the Proof of Relative Openness Theorem in Rudin's Principles of Mathematical Analysis

The subspace topology is usually defined as the statement given in the proposition. Rudin however defines "$E$ open relative to $Y$" in the context of metric spaces. The definition provided ...
K. Jiang's user avatar
  • 8,269
1 vote

Understanding the Proof of Relative Openness Theorem in Rudin's Principles of Mathematical Analysis

First off, it's important to understand what's being proved here is precisely that the "subspace topology" is $\textit{equal}$ to Rudin's definition of being "relatively open" (e.g....
tiny's user avatar
  • 70
1 vote

Manhattan metric proof

The idea behind your proof is fine, but you can adjust some of the wording to avoid confusion. For instance, do not start with your equation (1); this is what you are trying to prove. Start by ...
K. Jiang's user avatar
  • 8,269

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