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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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## 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).