## functions – \$ f: A rightarrow C \$ and \$ g: B rightarrow C \$, try \$ f cup g: A cup B rightarrow C \$ iff \$ f restriction (A cap B) = g restriction (A cap B) \$

Suppose $$f: A rightarrow C$$ Y $$g: B rightarrow C$$.

Test it $$f cup g: A cup B rightarrow C$$ iff $$f restriction (A cap B) = g restriction (A cap B)$$

My attempt:

If I get the restriction definition correctly, we have

$$f restriction (A cap B) = f cap bigl ((A cap B) times C bigr)$$

Y

$$g restriction (A cap B) = g cap bigl ((A cap B) times C bigr)$$

$$( rightarrow)$$

Suppose $$f cup g: A cup B rightarrow C$$

Drink $$(x, y) in f restriction (A cap B)$$. So $$(x, y) in (A cap B) times C$$ Y $$x in B$$.

As $$x in B$$, is there any $$b en B$$ such that $$(x, b) in g$$.

As $$(x, b) in f cup g$$ Y $$(x, y) in f cup g$$, we conclude that $$y = b$$. Thus $$(x, y) en g constraint A cap B$$Y $$f restriction (A cap B) subseteq g restriction (A cap B)$$.

By the same reasoning, we can show that $$g restriction (A cap B) subseteq f restriction (A cap B)$$

$$( leftarrow)$$

Suppose $$f restriction (A cap B) = g restriction (A cap B)$$

Existence: take $$x in A cup B$$. So $$x in A$$ or $$x in B$$, which implies that there is some $$c in C$$ such that $$(x, c) in f$$ or $$(x, c) in g$$. Thus $$(x, c) in f cup g$$.

Singularity: suppose $$(x, y), (x, p) in f cup g$$. If both elements are in $$f$$ or in $$g$$then clearly $$y = p$$.

Suppose $$(x, y) in f$$ Y $$(x, p) in g$$.

As $$x in A cap B$$ Y $$and in C$$, we have $$(x, y) in (A cap B) times C$$ and therefore $$(x, y) en f constraint A cap B$$

For similar reasoning, we have $$(x, p) en g constraint A cap B$$.

For all $$x$$ in $$Dom (f restriction A cap B)$$ there is only one $$k en Ran (f restriction A cap B)$$ such that $$(x, k) en f constraint A cap B$$

The same reasoning applies to $$g restriction A cap B$$.

As $$f restriction (A cap B) = g restriction (A cap B)$$, we have $$p = y$$.

Yes $$(x, p) in f$$ Y $$(x, y) in g$$, then by the same reasoning, we conclude that $$p = y$$

We have demonstrated existence and uniqueness, hence the result.

$$f cup g: A cup B rightarrow C$$
$$Box$$

It is right?

## 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?

## 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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## 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.