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# Tag: Interior

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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.

One more thing, when your homepage has a higher rating than your internal pages, you will notice that the results of your search will become unstable.

## 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:

- 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 $.
- 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 $.
- 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) $. - 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).

## $ 80 for the best barn doors and interior business logo

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## which is the best course of interior design in jaipur

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