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10

Strictly speaking, in order for $f\circ g$ and $g\circ f$ to both be defined and equal, then yes, $f$ and $g$ must both be functions $f:X\to X$ and $g:X\to X$ for some set $X$. (However, some people would allow the composite $k\circ h$ of functions $h:A\to B$ and $k:Y\to Z$ as long as $B\subseteq Y$, and are less strict about what it means for two functions ...


9

From the perspective of an elementary calculus student (by which I mean that it can supposedly be made rigorous later on, but isn't in introductory classes), the $du$, $dx$ stuff is absolute nonsense and I will never understand why it continues to be used by so many professors. It really is the one glaring hole in most otherwise rigorous calculus courses. ...


8

Technically, yes. The domain is a singleton, so the only topology on it is the trivial one, $\tau_0$. Then if $(Y,\tau)$ is any other topological space and $*$ is the one-point space $$f:(*,\tau_0)\to (Y,\tau)$$ is trivially continuous because if $U\in\tau$ is open $$f^{-1}(U)\in\{\varnothing, *\}=\tau_0$$ depending on whether or not $f(*)\in U$ or not. ...


7

This is an infinite string of digits. But this string, when we try to interpret it in decimals, does not encode a number, i.e., an element of ${\mathbb N}$ or ${\mathbb R}$. Natural numbers, when written in decimals, appear as finite strings, and have a last digit. Noninteger real numbers, when written in decimals, require a decimal point, after which an ...


6

It means that $A\subseteq B$ and that $f(a)=a$ for every $a$ in $A$. But $f$ is not the identity map unless $B=A$.


6

The empty set is a subset of every set, but it is not an element of every set. In your examples, $$\varnothing \in \{\varnothing, \{a\}\}\text{ but } \varnothing \notin \{\{a\}\},$$ but it is a subset of both sets: $$\varnothing \subset \{\varnothing, \{a\}\}, \text{ and } \varnothing \subset \{\{a\}\}.$$


5

This relation is transitive. 3 doesn't play any role here since you don't require a transitive relation to be full. Observe that the transitivity here means: $$(1,2)\wedge(2,1)\in R\Rightarrow(1,1)\in R \\(2,1)\wedge(1,2)\in R\Rightarrow(2,2)\in R$$ and these are found in the relation so it's transitive. In fact that's equivalence relation on $\{1,2\}$ ...


5

I suspect that there is no universal agreement among different sources. But for example Rudin's Principles (p. 94) says "If $x$ is a point in the domain of the function $f$ at which $f$ is not continuous, we say that $f$ is discontinuous at $x$, or that $f$ has a discontinuity at $x$". He doesn't mention anything about points not in the domain of $f$, but ...


5

They need not be bijections. $g(x)=x$ and $f(x)=0$ for all $x$. Then $f \circ g =0 = g \circ f$. $f$ is as far from a bijection as you will get.


5

I have studied Mathematics in the UK for many years, and I have never heard anyone saying that the first quadrant is the top-left region. The convention is what you say; where $x$ and $y$ are positive.


3

For a Banach space $X$ , the dual space is the Banach space $X^*$ of all bounded linear functionals on $X$. Let $\langle x,f\rangle$ be the value of $f\in X^*$ on $x\in X$. If $T$ is a bounded linear operator on $X$, then its adjoint is a bounded linear operator $T^*$ on $X^*$ which is defined by $\langle x,T^*f\rangle=\langle Tx,f\rangle$ for all $x\in X$ ...


3

Let $X$ be a non-empty set and $f$, $g:X \to X$ mappings. If $f = g$ is arbitrary, then $f \circ g = g \circ f$. If $g$ is the identity mapping, then $f \circ g = g \circ f$ for every $f$. If there exists an $x_{0}$ in $X$ such that $g(x) = x_{0}$ for all $x$, then $f \circ g = g \circ f$ for every $f$ that fixes $x_{0}$. Etc., etc. As others have noted, ...


3

If $f$ and $g$ are in the same path component of $\operatorname{Imm}(M, N)$, then there is a path $p : [0, 1] \to \operatorname{Imm}(M, N)$ such that $p(0) = f$ and $p(1) = g$; i.e. $p$ is a path connecting $f$ and $g$. Now consider $H : M \times [0, 1] \to N$ given by $H(x, t) = p(t)(x)$, then $H(x, 0) = p(0)(x) = f(x)$ and $H(x, 1) = p(1)(x) = g(x)$ and ...


3

I think the best way to understand such questions is that the asker has never properly internalized what "$=$" means. (They have probably been told, but it didn't catch hold). These askers appear to understand $A=B$ as meaning "$A$ becomes $B$, through some unspecified process". This understanding makes enough sense in many cases -- namely when $A$ is an ...


3

First of all, before we can callthis an integer, we can only call it a blob of ink or a bunch of pixels. How are some buches of pixels integers and others are not? Well, the are not integers, they represent integers. For example MMXV, and $2015$, and 0x7DF are distinct representations of the same integer, and all given with enough "context" to allow us to ...


2

According to your link a submatrix is just a block of another matrix. So this is just a connected 'rectangle' of numbers of the original matrix, that means (as you already assumed) that you cannot skip rows or columns. But if you consider following link, you can skip rows or columns, that means you can construct any submatrix by deleting whole columns and ...


2

Let $V$ be a complex vector space. A complex conjugation on $V$ is an antilinear map $\sigma : V \to V$ such that $\sigma\circ\sigma = \operatorname{id}_V$. Not every complex vector space comes with a complex conjugation, in general it is an extra piece of data. Complex Euclidean space $\mathbb{C}^n$ has a complex conjugation given by $\sigma(z_1, \dots, ...


2

Consider $\prod \limits_{i \in I} M_i$. The sum is $$( a_i)_{i \in I } + ( b_i)_{i \in I } = ( a_i + b_i)_{i \in I }$$ The action of R : $$ r \cdot ( a_i)_{i \in I } = ( r \cdot a_i)_{i \in I }$$


2

This baffled me too as I first came along, Jack M gave a good answer what really is behind it. And for the practicing mathematician, the „symbolic approach“ via variables, differentials and so on is just like a mental shorthand, that works by clever choosen notation. Maybe bear in mind that the chain rule could be read in two directions, one if you see at ...


2

As has been discussed in the comments, the argument of $0$ is taken to be either zero or undefined. It is a matter of convention.


2

Your proof of the first part is lacking as well. You only show that there is an example for the transitivity. You don't show that for every two pairs $(a,b)$ and $(b,c)$ also $(a,c)\in R$. Similarly for reflexivity and symmetry. So you did not, in fact, verify that this is an equivalence relation. As for the second part, given an equivalence relation, the ...


2

The sequence 2, 2, 2, 3, 4 is not increasing, but it's non-decreasing. The sequence 2, 3, 4, 5, 6 is both increasing and non-decreasing. Compare the difference between "less than" and "not greater than." (I have never seen the words ascending/descending used in this context but I imagine it's the same).


2

I've usually seen the question at the end written as: $$ 5 = \; ? $$ If you take the $=$ symbol literally at its usual meaning in these puzzles, then the puzzle statement is equivalent to solving for $x$ in $$ (\mbox{false} \wedge\mbox{false}\wedge\mbox{false}\wedge\mbox{false}) \Rightarrow (5 = x)$$ which is vacuously satisfied by any value of $x$, ...


2

I would call it a cylinder — a right cylinder if the movement is normal to the plane of the original two-dimensional region. However, cylinder is used in enough different senses that I’d probably explain my usage first.


2

Identifying the graph with the function is a common approach in set theory, especially elementary set theory. It is awkward in other disciplines, where it is more convenient for a function to be a special kind of object which can have properties that are specific to functions (continuity, for example). A case of this from topology is that if $A$ and $B$ ...


2

For what it's worth here is a characterisation of the finite semigroups that are regular with zero in which the product of any two different idempotents equals zero. If $S$ and $T$ are semigroups with zero, then the 0-union of $S$ and $T$ is the semigroup formed by taking the union of $S$ and $T$ but identifying their zero elements, and with multiplication ...


2

For a function $f$ to be continuous at a point $a$, you must have $a\in\text{dom}(f)$. The function you cite is continuous on the punctured plane.


2

Start with a point $x$. Let $I$ be an open interval containing $x$. The oscillation of $f$ on $I$ is the quantity $\displaystyle \sup_{s,t \in I} |f(t) - f(s)|$. For all such $I$ containing $x$ you get a value for the oscillation of $f$ on $I$. The oscillation of $f$ at the point $x$ is the infimum of all such values.


2

What happens if $\varnothing$ is in fact an element of some set? For example, $\{\varnothing\}$? How can you tell the difference between $\varnothing$ and $\{\varnothing\}$, if you add the empty set as an element to every set? Membership and inclusion are two different relations, and should be treated differently.



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