Private Proxies – Buy Cheap Private Elite USA Proxy + 50% Discount!Private Proxies – Buy Cheap Private Elite USA Proxy + 50% Discount!Private Proxies – Buy Cheap Private Elite USA Proxy + 50% Discount!Private Proxies – Buy Cheap Private Elite USA Proxy + 50% Discount!
    0
  •   was successfully added to your cart.
  • Home
  • Buy proxies
  • Extra features
  • Help
  • Contact
  • Login
  • 50% OFF
    BUY NOW!
    50
    PROXIES
    $19
    --------------------
    BUY NOW!
    BUY NOW!
    BUY NOW!
    BUY NOW!
    BUY NOW!
    $29
    $49
    $109
    $179
    $299
    --------------------
    --------------------
    --------------------
    --------------------
    --------------------
    PROXIES
    PROXIES
    PROXIES
    PROXIES
    PROXIES
    100
    200
    500
    1,000
    2,000
    TOP SELLER
    BEST VALUE
    For All Private Proxies!

Here is a question I asked myself years ago. Since it is not really in my field, I hope to find some (partial) answers here…

The chromatic number of a graph $ G$ is the minimal cardinal of a partition of $ G$ into independant sets, i.e. sets all of whose connected components are trivial. This definition of independant sets suggests that a good analogue of them in the setting of topological spaces would be totally disconected sets. This motivate the following definition:

Definition. Define the chromatic number of a topological space $ X$ , denoted by $ \chi(X)$ , as the minimal cardinal of a partition of $ X$ into totally disconnected sets.

My main question is the following:

Question. What is the chromatic number of $ \mathbb{R}^n$ ?

Some remarks. If $ X$ can be embedded in $ Y$ , then $ \chi(X) \leqslant \chi(Y)$ . We also have that $ \chi(X \times Y) \leqslant \chi(X)\chi(Y)$ , because if $ (A_i)_{i \in I}$ and $ (B_j)_{j \in J}$ are respective partitions of $ X$ and $ Y$ into totally disconnected sets, then $ (A_i \times B_j)_{(i, j) \in I \times J}$ is so for $ X \times Y$ . We also have that $ \chi(\mathbb{R}) = 2$ , witnessed by the partition $ \{\mathbb{Q}, \mathbb{I}\}$ , where $ \mathbb{I}$ denote the set of irrational numbers. So $ \chi(\mathbb{R}^n) \leqslant 2^n$ .

But we have better. Actually, $ \chi(\mathbb{R}^2) \leqslant 3$ , witnessed by the partition $ \{\mathbb{Q}^2, (\mathbb{Q} \times \mathbb{I} \cup \mathbb{I}\times \mathbb{Q}), \mathbb{I}^2\}$ (where $ \mathbb{Q} \times \mathbb{I} \cup \mathbb{I}\times \mathbb{Q}$ is totally disconnected because it embeds into $ \mathbb{I}^2$ via $ (x, y) \mapsto (x + y, x – y)$ ). So $ \chi(\mathbb{R}^{2n}) \leqslant 3^n$ and $ \chi(\mathbb{R}^{2n + 1}) \leqslant 2 \times 3^n$ .

My conjecture is that $ \chi(\mathbb{R}^n) = n + 1$ . For the upper bound, I suspect that some partition of the form $ \{A_0, \ldots A_n\}$ where $ A_i$ is the set of elements of $ \mathbb{R}^n$ having exactly $ i$ rational coordinates could work, but I don’t manage to prove anything. For the lower bound, I only have an intuition given by the following “image”: if you consider $ A \subseteq \mathbb{R}^n$ totally disconnected, then it sounds reasonable to think that there exists a set $ B\subseteq \mathbb{R}^n$ , homeomorphic to $ \mathbb{R}^{n – 1}$ , passing “between” the points of $ A$ ; however, this is just an inutuition and I don’t know if it is true. I suspect that some algebraic topology would be needed to prove that, but I have almost no experience in this field so I couldn’t investigate more; I am not even able to show that $ \chi(\mathbb{R}^n) > 2$ for some $ n$ .

An other possibility would be that $ \chi(\mathbb{R}^n) = 2$ for every $ n$ , because of some “weird” colouring. What makes me suspect that is that if we replace “connected” by “arcwise connected” in the definition of the chromatic number, then it becomes easy to build a partition of $ \mathbb{R}^n$ into two parts each of which having no non-trivial arcwise connected subset, by a diagonal argument using the axiom of choice. However, this proof uses the fact that there are exactly $ \mathfrak{c}$ arcs $ \gamma : [0, 1] \longrightarrow \mathbb{R}^n$ . To do the same proof for connected sets, we would need the existence of a family of $ \mathfrak{c}$ non-trivial connected subsets of $ \mathbb{R}^n$ such that every non-trivial connected subset of $ \mathbb{R}^n$ has a subset in this family, and I don’t think that such a family exists (but I don’t know).

In case where we manage to prove that $ \chi(\mathbb{R}^n) = 2$ thanks to some construction using the axiom of choice, it would be interesting to investigate what happens if we impose some restriction on the complexity of the sets in the partition, for example to be Borel.

✓ Extra quality

ExtraProxies brings the best proxy quality for you with our private and reliable proxies

✓ Extra anonymity

Top level of anonymity and 100% safe proxies – this is what you get with every proxy package

✓ Extra speed

1,ooo mb/s proxy servers speed – we are way better than others – just enjoy our proxies!

50 proxies

$19/month

50% DISCOUNT!
$0.38 per proxy
✓ Private
✓ Elite
✓ Anonymous
Buy now

100 proxies

$29/month

50% DISCOUNT!
$0.29 per proxy
✓ Private
✓ Elite
✓ Anonymous
Buy now

200 proxies

$49/month

50% DISCOUNT!
$0.25 per proxy
✓ Private
✓ Elite
✓ Anonymous
Buy now

500 proxies

$109/month

50% DISCOUNT!
$0.22 per proxy
✓ Private
✓ Elite
✓ Anonymous
Buy now

1,000 proxies

$179/month

50% DISCOUNT!
$0.18 per proxy
✓ Private
✓ Elite
✓ Anonymous
Buy now

2,000 proxies

$299/month

50% DISCOUNT!
$0.15 per proxy
✓ Private
✓ Elite
✓ Anonymous
Buy now

USA proxy location

We offer premium quality USA private proxies – the most essential proxies you can ever want from USA

100% anonymous

Our proxies have TOP level of anonymity + Elite quality, so you are always safe and secure with your proxies

Unlimited bandwidth

Use your proxies as much as you want – we have no limits for data transfer and bandwidth, unlimited usage!

Superfast speed

Superb fast proxy servers with 1,000 mb/s speed – sit back and enjoy your lightning fast private proxies!

99,9% servers uptime

Alive and working proxies all the time – we are taking care of our servers so you can use them without any problems

No usage restrictions

You have freedom to use your proxies with every software, browser or website you want without restrictions

Perfect for SEO

We are 100% friendly with all SEO tasks as well as internet marketing – feel the power with our proxies

Big discounts

Buy more proxies and get better price – we offer various proxy packages with great deals and discounts

Premium support

We are working 24/7 to bring the best proxy experience for you – we are glad to help and assist you!

Satisfaction guarantee

24/7 premium support, free proxy activation and 100% safe payments! Best reliability private proxies for your needs!

Best Proxy Packs

  • 2,000 Private Proxies $600.00 $299.00 / month
  • 1,000 Private Proxies $360.00 $179.00 / month

Quick Links

  • More information
  • Contact us
  • Privacy Policy
  • Terms and Conditions

Like And Follow Us


Copyright ExtraProxies.com | All Rights Reserved.
  • Checkout
  • Contact
  • Help
  • Home
  • My Account
  • My Cart
  • News
  • Privacy Policy
  • Proxy features
  • Proxy packs
  • Terms and Conditions
Private Proxies – Buy Cheap Private Elite USA Proxy + 50% Discount!
    0 items