real analysis: shows that $ m ( {x in [0, pi/2] The | cos (x) in mathbb {Q} }) = m ( { [0, pi/2] cap mathbb {Q} }) = 0. $ (if the statement is correct)

Showing that $ m ( {x in (0, pi / 2) | cos (x) in mathbb {Q} }) = m ( {(0, pi / 2) cap mathbb { Q} }) = 0. $

My questions are:

to try $ subseteq $:

Assume that $ 0 leq x leq pi / 2 $, then taking the cosine of this, we get $ cos 0 geq cos x geq cos ( pi / 2), $ so, $ 1 geq cos (x) geq 0 $ (why $ cos (x) $ it is a decreasing function in this interval.) ….. is my test correct?

to try $ supseteq $:

I think this address is not true, I think the correct statement should be $ m ( {x in (0, pi / 2) | cos (x) in mathbb {Q} }) subseteq m ( {(0, pi / 2) cap mathbb {Q} }) = 0. $ I am right?

Could someone help me answer those questions, please?

pathfinder 1e – What is the cap of the mythical shield of faith?

Under the mythical rules, there are mythical spells. The mythical shield of faith adds your level to the deviation bonus of the base spell, but does that mean you reach the limit bonus provided much earlier or is that level bonus in addition to the spell bonus?

So, if I was a pitcher with a level 3 pitcher level and threw a shield of mythical faith, is the diversion bonus +5 or +8?

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linear algebra: find bases for $ S cap T, S $ and $ T $. $ S + T = V? $

Leave $ V = M_ {2,3} $($ mathbb C $), $ hspace {1mm} $leave $ S = Bigg {A en V: A $$ begin {pmatrix} 1 \ 2 \ – 1 end {pmatrix} = begin {pmatrix} 0 \ 0 \ end {pmatrix} $$ Bigg } $

Y $$ T = big {A en V: left (1 hspace {3mm} 2 right) A = left (0 hspace {3mm} 0 hspace {3mm} 0 right) big } $$ We can assume that S and T are subspaces of V.

I) Find bases for $ S cap T, S $ Y $ T $.

ii) Make $ S + T = V? $

I'm not sure how to do the $ S cap T $ part? Also by ii) I know that $ dim (S cap T) -dim (S) -dim (T) = dim (S + T) $ cast $ dim (S + T) leq dim (V) = 6. $ Then I could deduce the answer if I knew $ S cap T $. Let me know if there is a simpler method.

analysis – Limit of $ A_j cap B_j $ on the distance from Hausdorff, when $ A_j rightarrow A $ and $ B_j rightarrow B $.

Leave $ A_j $, $ B_j $, $ A $ Y $ B $ is set to $ mathbb {R} ^ n $. Leave $ A_j rightarrow A $ Y $ B_j rightarrow B $ both in the Hausdorff distance (induced by the usual Euclidean metric in $ mathbb {R} ^ n $) Y $ A_j cap B_j neq emptyset $ for all $ j in mathbb {N} $. Yes $ A_j cap B_j rightarrow E $ for some set $ E subset mathbb {R} ^ n $so, we have $ E subset A cap B $?

It's known that $ E neq A cap B $ in general, but the counterexample that I could only find falls in the case $ E subset A cap B $. So is it true in general?

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Camera: Fujifilm Instax mini 90 lens cap does not close

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Why $ v in C ^ 2 (D) cap C ( overline {D}) $ and $ Delta v = – lambda v $ in $ D $ implies $ v in C ^ 2 ( overline {D}) $

The real problem is to show that the problem pde, where $ v = 0 $ in $ partial D $ plus the previous hypothesis, it has a non-trivial solution only if $ lambda geq 0 $. Which is pretty trivial if you use one of Green's identities. But for that it is necessary that $ v in C ^ 2 ( overline {D}) $, and I do not see why that is true.

How can I know if this "2 in 1" lens cover and cap will be effective on my lens?

… my wife has the (bad) habit of putting her keys next to my glasses and my three-year-old son recently acquired a new hobby. Collecting rocks.

Some options to consider:

  • Use magnifying rings so you can use the same lens caps with all your lenses. This would prevent you from having any loose cover to lose.

  • Use neoprene bags. While they may be additional items to lose, the risk is reduced if you bring only enough to cover lenses that are not connected to your camera.

  • Place a small drawstring bag in the stroller to place various items, such as lens caps and your spouse's keys.

  • Encourage your spouse to carry a bag in which you can place your keys. Children may also need to start carrying their own bags or backpacks. They are only allowed to take rocks that they can carry themselves.

  • Take only one lens with you, connected to the camera. You are at the center of the target market for zoom trips.

  • Buy caps in bulk, so it does not matter if you lose some.

… [the hood] It does not make any difference. I still keep the hood on as protection. – andreas

If the hoods do not make any difference, it will not matter which one you wear. As long as the vignetting is not introduced, you will not notice any difference in the quality of the image, that is what your question is about. However, a shorter hood would offer less "protection", which would be compensated with the built-in lid. The only way to know if it is adequate for your needs is to try it.

How much difference will those mm make in the image quality?

The only way to know is to try it.The parasols of the effect lenses have depends on the specific lenses with which you use them. For example, lenses with good coatings benefit less than lenses with poor coatings. Since Sigma Art lenses seem to be well seen, I would expect their coatings to be good and the hoods to be less beneficial.

To test your lens, find a scene that you think would benefit from a hood. Photograph it with your camera on a tripod, with and without a hood. This will allow you to know the range of results you can expect. If you can not tell the difference, the hood is unnecessary.

I do not even bother with the hoods of my best lenses because I do not see any difference in the test images taken in difficult conditions. I also do not bother with the covers because I have found that images taken with lens caps are not acceptable.