# Tag Info

10

No; take the two-point topological space with one open point and one closed point. The open point is compact, but not closed. In fact, I don't think that there are any topological spaces which are closed inside every other topological space.

9

Great question! The answer is yes if you restrict your attention to sufficiently nice spaces. Specifically, each of the following statements is true (1 and 3 are quite well-known): Let $Y$ be a Hausdorff space and $X\subseteq Y$ be a compact subspace. Then $X$ is closed in $Y$. Let $X$ be a regular space such that whenever $X$ is a subspace of a ...

6

Formally speaking, neither is correct. Sequences are functions, and the elements of $M$ are not [usually] functions from $\Bbb N$ to $M$ itself (although that is known to be possible). What is correct is that "$(x_n)$ is a sequence such that $x_n\in M$ [for all $n$]". To me, then, both notations abuse about the same the correct way of phrasing, so both are ...

6

Neither $(x_n)\subset M$ nor $(x_n)\in M$ should be used, since neither is correct. The fact that we say "sequence in $M$" does not mean we should use the symbol for element in a set. There are clear and precise definition for the meaning of $\in$ and $\subset$ and they dictate what the correct usage is. Generally speaking, a sequence $(x_n)$ of elements in ...

4

Division by zero is undefined in every case. In calculus, the phrase “$0/0$ is an indeterminate form” means that you have a limit of the form $$\lim_{x\to a}\frac{f(x)}{g(x)}$$ where $$\lim_{x\to a}f(x)=0 \qquad\text{and}\qquad \lim_{x\to a}g(x)=0$$ but $f(x)/g(x)$ is defined in a set having $a$ as a limit point (usually, but not necessarily, a ...

3

$x/y=z$ means that $zy=x$, i.e. $x/y$ is the number you multiply by $y$ to get $x$. For $1/0$, there is NO number you can multiply by zero to get 1. For $0/0$, there is no UNIQUE number you can multiply by zero to get 0.

2

"Replace" is a strong word. The triangle inequality in normed spaces $(X,\|\cdot\|)$: $$\|x+y\| \leq \|x\|+\|y\|, \quad \forall\,x,y \in X$$ is more general, in the sense that it can take a lot of forms depending on which space and norm you're dealing with, for example: \left(\sum_{n \geq 1} |x_n+y_n|^p\right)^{1/p} \leq \left(\sum_{n \geq 1} ...

1

Without looking too deeply into it, I would first propose that undefined is more along the lines of not solvable. For ex: x/0 is not solvable until we know the definition of x. Undetermined is not yet solved. 0/0 is not yet solved, but we can use L'Hopital's rule to (often) eventually solve it. (I'd like to caveat that this is an initial impression, so make ...

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