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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.
Mar
17
comment Is $\prod \limits_{i = 1}^{n} [0,1] \subseteq \mathbb R^n$ homeomorphic to the closed unit ball?
This proof can be extended (but it's more details to check) to show that every closed and bounded convex subset of $\mathbb{R}^n$ that has non-empty interior is homeomorphic to the unit ball in that space. The idea is the same: from an interior point, extend rays emanating from that point, all of which lie inside the convex body, up to some unique boundary point. Then map all these segments to the corresponding one in the unit ball wrt $0$.
Mar
14
comment Every order topology is regular (proof check)
The last bit is a typo: it should say $V = V_1 \cap V_2$. $U = U_1 \cup U_2$ is correct. The extra checks are needed because the point $x$ could have a direct neighbour on either side, which necessitates a more cumbersome notation. Consider $x = 1$ and $A = \{2,3\}$ in the natural numbers, in the order topology e.g.
Mar
14
comment Every order topology is regular (proof check)
regular here means $T_1$ plus every closed set and point can be separated by disjoint open sets (so $T_3$). So $T_1$-ness, or even $T_2$-ness have to be shown as well. It's not used in the second part, but it is necessary for the property as defined in Munkres.
Mar
14
answered Existence of a continuous bijection in the plane
Mar
8
comment Neighborhoods and Metric Spaces in Real Analysis
2 is just a reformulation of 1. All $x$ with $\rho(x,a) < \epsilon$ is exactly the set $B_\epsilon(a)$.
Mar
8
comment Prob. 5 in Exercises after Sec. 17 in Munkres' TOPOLOGY, 2nd ed.: How to prove this result in a general ordered set?
math.stackexchange.com/a/697946/4280 has my answer..
Mar
8
answered Topology and “adding sets” in a TVS
Mar
8
answered Examples of Separable Spaces that are not Second-Countable
Mar
7
answered Problem book on general topology
Mar
7
comment Are the rationals minus a point homeomorphic to the rationals?
@GrumpyParsnip Yes, that's the one. It's from the same era as the ones for the Cantor set (due to Brouwer) e.g. There are also characterisations for other spaces like $\mathbb{Q} \times \mathbb{P}$ etc. See Fons van Engelen's thesis "Homogeneous Zero-Dimensional Absolute Borel Sets", e.g.
Mar
6
answered Additional properties of closure