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6h
comment A subspace of a metric space is normal
The Hausdorff axiom does not imply that every two closed sets can be separated by disjoint open sets. In other words, "Hausdorff" does not imply "normal".
6h
comment A subspace of a metric space is normal
@Mathemagician1234 That all metric spaces are normal is a standard theorem. To prove it, first check that, for any closed set $A$, the distance function $x\mapsto d(x,A)$ is continuous, and then use the open neighborhoods $\{x:d(x,A)<d(xB)\}$ and $\{x:d(x,A)>d(xB)\}$ to separate $A$ and $B$.
2d
comment A subspace of a metric space is normal
Your argument is correct. Every subspace of a metric space is itself a metric space and is therefore normal.
2d
comment A subspace of a metric space is normal
Normality requires disjoint open neighborhoods for each pair of disjoint closed sets, not just for disjoint closed balls of the same radius. Furthermore, the $A$ and $B$ in your answer are not necessarily disjoint. In fact, the only time they're disjoint is if the two balls $B_r(x)$ and $B_r(y)$ cover the whole space $X$.
2d
comment Homotopy of a pair
@user135988 No, $A=X$ only in Question 2, not in Question 1. Also, it's not clear what you're asking for in Question 1, since you already gave the most natural answer: You defined $\pi_n(X,A,x)$, so just specialize to $n=1$ (and $x=x_0$).
2d
comment Definition of convergence in the product and in the box topology
I don't think the OP's formulation of the definition of convergence is correct. It should not have $\forall x\in X$, and $N(x)$ should be simply $N$.
2d
comment Are the any **non-trivial** functions where $f(x)=f'(x)$ not of the form $Ae^x$
@MichaelHardy You're quire right that the use of "any" can lead to ambiguities. In Robert Israel's comment, though, it does not; the comment is unambiguous and his use of "any" is fine.
2d
comment Generalization of Burnside theorem
A proof is not a question.
2d
comment Presentation of a group question
I'd disagree with the part of your first sentence where you say that you can derive the group elements form the relations in any group presentation. The elements of $G$ are given already by the generators in the presentation, not the relations. The elements of $G$ are all the products of generators and their inverses in all possible arrangements. What the relations tell you is which of these products are equal to each other, so you can trim down the list of elements by removing duplicates.
May
19
comment Representing linear transformations as matrices. What benefit, if any, is there to *not* expressing them as diagonal matrices?
This is an excellent answer to the title, but the question itself shows that the OP is confused about something much more basic.
May
19
answered mathematical symbol for vector appending
May
19
comment Lambda calculus typing
How about $(A\to(A\to B))\to(A\to B)$, for arbitrary types $A$ and $B$?
May
18
comment Countable Product of Sequentially Compact spaces is Sequentially Compact
@ChrisMarais Maybe I should point out explicitly that, even though the process involves infinitely many promises, at any particular stage of the process, when you're thinning out the sequence, you have only finitely many prior promises to keep; most of the infinitely many promises are yet to come.
May
18
answered Indiscernibles and colorations.
May
18
comment Condition for an Ultrafilter to be Ramsey.
In your "initial idea", you considered assigning colors to elements of $\omega$ (as each $A_n$ is a subset of $\omega$). You should be assigning colors to two-element subsets of $\omega$ because the domain of $c$ is $[X]^2$, not $X$. Also, note that you've misquoted the definition; it should be an element of $D$, not of $X$, that is homogeneous.
May
18
answered Countable Product of Sequentially Compact spaces is Sequentially Compact
May
17
answered A function symbol with more than one arity schema and type assigned to it in a signature
May
16
comment Must $\mathfrak{sd} = \mathfrak{d}$?
Although I'm responsible for a few multi-character names of cardinal characteristics, I would advise against combinations like $\mathfrak{sd}$ that can be easily read as the product of $\mathfrak{s}$ and $\mathfrak{d}$ (thereby making the question really easy.)
May
16
comment Must $\mathfrak{sd} = \mathfrak{d}$?
In "apply $s_i$ to all those finite sequences, I was tacitly thinking of $s_i$, by virtue of your requirement 2, as taking finite sequences of length $k-1$ as inputs and producing sequences of length $k$ as outputs, and in particular producing as outputs the last components of those sequences.
May
16
answered Must $\mathfrak{sd} = \mathfrak{d}$?