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general topology – Can the interior of a smooth (not empty) collector be empty with a limit?

Leave $ M $ Be a smooth collector with limit.

Yes int$ M = emptyset $, so $ M = partial M $ it would be a mild variety without limit, and so every point of $ M $ It is an interior point. So we have $ M = $In t$ M = emptyset $.

Is my argument correct? In other words, if I have a flat non-empty collector $ M $ with limit, I am authorized to assume Int.$ M ne emptyset $?

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Interior pages Vs Homepage

Google is more concerned about the internal pages of a site than its home page because the information that will be useful to search for users is often not found on the home page.

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gn. general topology: closed set that is the closing of its interior points.

This is a cross-publication of MSE for this question.

https://math.stackexchange.com/questions/3267497/a-set-which-is-the-closure-of-its-interior-points

There is a related question in MO:
Closing the inside of a convex set in a topological vector space. Unfortunately, in my problem, I do not have the convexity and, however, the convexity and the softness of the limit are quite different assumptions (a priori, the closed set with the soft limit can be non-convex). Here is the question.

I'm trying to give a sufficient condition for a set in $ mathbb {R} ^ n $ What is the closure of your interior points. A priori, such a set has to be a closed set. A closed set in general is not the closure of its interior point. An example of a trivial counter is a set with an empty interior, for example, the segment $[0,1] times {0 } $ in $ mathbb {R} ^ 2 $. So the next candidate is one with the interior not empty. An example of a counter can be a sphere with hair, for example, the union of $ {x ^ 2 + y ^ 2 le 1 } cup [1,2] times {0 } $ in $ mathbb {R} ^ 2 $. So to avoid this kind of counter example, I propose the following question (can it be "conjecture"?)

Question 1: Let's go $ K $ be a closed connected subset of $ mathbb {R} ^ n, n ge 2 $ with interior not empty $ Int (K) $. Suppose that the limit $ partial K $ it's a connected submanifold of $ mathbb {R} ^ n $. So $ K = overline {Int (K)} $.

Note: by submanifold of $ mathbb {R} ^ n $ I mean an embedded submanifold of $ mathbb {R} ^ n $. In particular, $ partial K $ It is orientable.

The assumption of connectivity in $ K $ is to avoid the case that $ K $ It has an isolated point. I'm not sure I need the limit connectivity. $ partial K $. The reason why I put it there because the fact that $ K $ is connected does not imply that $ partial K $ Are you connected. This is due to this question.
https://math.stackexchange.com/questions/170337/connectedness-of-the-boundary
what it says that (if I understood correctly) a connected set $ K $ has bound limit if and only if the add $ K ^ c = mathbb {R} ^ n setminus K $ Are you connected. Again, the opposite example to this situation (connected set with limit not connected) does not satisfy the supposition that the limit itself is a sub-matrix of $ mathbb {R} ^ n $. But even with this question I do not know how to approach yet, so let's just suppose that $ partial K $ it's a connected submanifold of $ mathbb {R} ^ n $.

My intent: I am able to give an argument in the case when $ K $ It is compact My argument is as follows:

  1. As $ K $ connected, $ Int (K) $ it has no isolated point and $ overline {Int (K)} = Int (K) cup partial Int (K) $. Note that $ K = Int (K) cup partial K $ A) Yes $ partial Int (K) subset partial K $.
  2. As $ K $ It's compact, $ partial K $ It is a compact and connected subcomposite of $ mathbb {R} ^ n $. We denote by $ k $ the codimension of $ partial K $. It's enough to prove that $ partial Int (K) = partial K $.
  3. Yes $ k ge 2 $: consider a point $ x in partial Int (K) $ and a local letter $ (U, varphi) $ of $ mathbb {R} ^ n $ so that $ varphi (U) = mathbb {R} ^ n $ Y
    $$ varphi (U cap partial K) = mathbb {R} ^ {n-k} times {0 } $$
    As $ k ge 2 $it turns out that $ U setminus partial K $ Are you connected. On the one hand, from $ x in partial Int (K) $, $ U cap Int (K) neq emptyset $. On the other hand, we have a decomposition.
    $$ U setminus partial K = (U cap Int (K)) cup (U cap K ^ c) $$
    of $ U $ As two disjointed open sets. So the connectivity of $ U setminus partial K $ implies that $ U cap K ^ c = emptyset $, that is to say $ U subset K $. As $ U $ open, $ U subset Int (K) $ what produces a contradiction from $ U $ contains a limit point of $ Int (K) $.
  4. We deduce that $ partial K $ It is a compact hypersurface of $ mathbb {R} ^ n $ which is orientable. By

"Lima, Elon L." The Jordan-Brouwer separation theorem for smooth hypersurfaces. "The American Mathematical Monthly 95.1 (1988): 39-42.",

I can deduce that $ mathbb {R} ^ n setminus partial K $ It has two components connected $ U_1, $ 2 whose limits are exactly $ partial K $. As $ K ^ c $ It is connected, without loss of generality, I assume that $ K ^ c subset U_1 $. It's enough to prove that $ Int (K) subset U_2 $from there $ Int (K) = $ U_2 and it follows that $ partial U_2 = partial Int (K) = partial K $.

Suppose that $ Int (K) setminus U_2 neq emptyset $A) Yes $ Int (K) cap U_1 neq emptyset $. Note that
$$ U_2 = (U_2 cap Int (K)) cup (U_2 cap partial Int (K)) cup (U_2 cap overline {Int (K)} ^ c). $$
Hence, the connectivity of $ U_2 $, we can deduce that $ U_2 cap partial Int (K) neq emptyset $ which is a contradiction since $ partial Int (K) subset Int (K) cap U_2 = emptyset $.

(This is, in fact, an argument in favor of a general fact: if $ U $ is an open open subset of $ mathbb {R} ^ n $ then for each $ V subset U $ Y $ V neq U $, we have $ partial V cap U neq emptyset $.)

Is my argument correct? I'm really grateful if someone can provide a proof or an opposite example of Question 1, with or without the connectivity hypothesis of $ partial K $. Thank you

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Image quality: How can I best use two interior flashlights to take pictures of my artwork indoors?

Response to the challenge of the box: On cloudy days, just leave. On sunny days, go to a patio or use a thin white sheet over a large window. Alternatively, you can build a scrim using PVC pipes and anything that diffuses the light (the sheets work well, I also built one with a more industrial version of parchment / waxed paper).

How to properly use 2 domestic torches to take Photo indoor ?: Use the same principles as the previous ones: put something thin in front of the torch to diffuse the light. However, two torches will be so dim that you will need to use a tripod and a very long exposure to avoid having to manipulate the ISO. Alternatively, experiment with light painting (however, this probably will not work well since you will end up having a "scratched" image.) This technique is not normally used to illuminate illustrations, but is used to create interesting images).

Do not be a cheap answer: Yongnuo manufactures some cheap speed lights that could be used, or you can try to find a used speed light in good condition. The rebound of these on the ceiling / wall will work well to diffuse the light. Studio Strobes can also be used often, or an entry level or similar AlienBee400 can also be purchased at a very low cost (in the grand scheme of things).

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