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May
23
comment How to prove that $(G,*)$ is a group?
You need to prove your claim that $G$ is closed under $\ast$, not just state it.
May
23
revised How to prove that $(G,*)$ is a group?
added 1 character in body
May
20
comment Definition of convergence in the product and in the box topology
@AndreasBlass You're right of course. I overlooked that.
May
20
comment Compactness and directed systems of subspaces
Both covers are similar in the sense that the last condition of the second formulation holds for both (they are inductive covers). The first condition for closed covers trivially holds for open covers (so these covers are "point-closure evading"). This might point to a generalisation: consider all directed, inductive covers with "point-closure evading" members.
May
20
answered If $\forall V\subseteq X$ where $x\in \overline V; f(x) \in \overline{f(V)}$, then $f$ is continous in $x$
May
20
comment If $\forall V\subseteq X$ where $x\in \overline V; f(x) \in \overline{f(V)}$, then $f$ is continous in $x$
What is $\color{red}{(B)}$ then exactly? And $f(x) = \overline{f(X\setminus f^{-1}(U))}$ makes no sense.
May
20
revised If $\forall V\subseteq X$ where $x\in \overline V; f(x) \in \overline{f(V)}$, then $f$ is continous in $x$
Latex
May
20
comment A question about the well-ordering theorem
I think you could assume that, yes. We could assume $J$ is a cardinal in its standard ordinal well-order, if you know some set theory. Then the same index set is "canonical" in a way, and the same one could be used for many sets. But taking the identity and using the promised well-order from the first theorem would work too.
May
20
revised A question about the well-ordering theorem
typos
May
20
comment Definition of convergence in the product and in the box topology
What is the topology of uniform convergence if not the one using the metric definition?
May
20
answered Definition of convergence in the product and in the box topology
Apr
15
answered If $g: Y \to Z$ is a continuous injection, then a map $f : X \to Y$ is open if $g\circ f$ is open.
Apr
15
answered $\{K∈K(X):K⊆U\}$ for $U$ open in $X$ generates $\textbf{B}(K(X))$
Apr
12
comment Given any two points $x, y \in X$ there exists connected set $A$ such that $x, y \in A$. Then $X$ is connected.
Another approach is to assume $X$ is not connected, so essentially your $f$ is onto, and deduce a contradiction from some $p$ with $f(p) = 1$ and $q$ with $f(q) = -1$. But that's essentially the same idea. Your proof is OK (with the one edit I made), and which @tattwamasiamrutam suggested.
Apr
12
revised Given any two points $x, y \in X$ there exists connected set $A$ such that $x, y \in A$. Then $X$ is connected.
edited body
Apr
11
comment Cardinality of an open dense set in a compact Hausdorff space
$\beta \mathbb{R}$ is also connected and is separable and has the same size as $\beta \mathbb{N}$
Apr
3
revised Lower limit topology doesn't arise from metric; proof by contradiction.
added 2 characters in body
Mar
25
revised Extension of a function from N to N to a continuous function in the Stone Cech compactification
added 5 characters in body
Mar
22
comment Projection map being a closed map
Suppose $U$ intersects $\pi[Z]$, say $x = \pi(x,y) \in U$, where $(x,y) \in Z$. Then $y \in V(y_i)$ for some $y_i$, and as $x \in U$, $x \in U(y_i)$ as well ($U$ is the intersection). But then $(x,y) \in (U(y_i) \times V(y_i)) \cap Z$, contrary to how to they were chosen disjoint from $Z$.
Mar
17
comment Affine cipher does not satisfy the diffusion property.
It will depend on how you define diffusion formally. So please state how you define it.