Henno Brandsma
Reputation
26,443
95/100 score
 Apr15 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. Apr15 answered $\{K∈K(X):K⊆U\}$ for $U$ open in $X$ generates $\textbf{B}(K(X))$ Apr12 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. Apr12 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 Apr11 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 $\beta \mathbb{N}$ Apr3 revised Lower limit topology doesn't arise from metric; proof by contradiction. added 2 characters in body Mar25 revised Extension of a function from N to N to a continuous function in the Stone Cech compactification added 5 characters in body Mar22 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 $Z$. Mar17 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. Mar17 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 $0$. Mar14 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. Mar14 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. Mar14 answered Existence of a continuous bijection in the plane Mar8 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 $B_\epsilon(a)$. Mar8 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.. Mar8 answered Topology and “adding sets” in a TVS Mar8 answered Examples of Separable Spaces that are not Second-Countable Mar7 answered Problem book on general topology Mar7 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. Mar6 answered Additional properties of closure