Search Results

Yes. Let Y be any compactly generated connected Hausdorff space such that Y fails to be locally compact. Let X be the one point compactification. X is maximal compact (since compact subsets of X are c …

The following canonical construction yields a `Yes' answer if $X$ is countable. Each $x \in X$ is assigned a metric space (or spaces) as follows. Let $U_{1},U_{2},...$ be a countable basis for $X$. …

Take [0,1] with the usual topology and attach a new point y so that the neighborhoods of y are precisely those which are and open dense in [0,1]. To obtain a 2nd disjoint copy, repeat the construction …

Replace [connected sets map to connected sets] with [point preimages are closed], and we have the following. Theorem. Suppose $X$ and $Y$ are compactly generated weakly Hausdorff spaces and suppose $ …

To create a counterexample X, start with the closed interval [0,1] (with the usual topology) and attach a new point z whose neighborhoods are open dense subsets of [0,1]. Observe [0,1] is a compact n …

The answer to the original question is apparently no. If a starting loop factors through a dendrite, and we delete a subloop which also factors through a dendrite, the surviving loop might not factor …

Definition 1. A dendrite $X$ is a 1 dimensional retract of the closed unit disk. Equivalently, $X$ is a compact, locally path connected, and uniquely arcwise connected metric space. Definition 2. The …

No. Let $X$ be a compactly generated Hausdorff space which fails to be locally compact at precisely one point. Now take the Alexandroff compactification of $X$, adding exactly one new point, whose nei …

Start with the one point compactification of the minimal uncountable well ordered space and then split the maximum point into two points.

No, even if $Y=[0,1]$. The piecewise linear continuous nondecreasing surjection $f:[0,1] \rightarrow [0,1]$ which maps $[1/3,2/3]$ to $1/2$ and is otherwise 1-1 and linear has no continuous section.

Suppose the space $X$ has a countable basis and $X$ is $T_{0}$. Must there exist a separable metrizable space $Y$ and a quotient map q:$Y \rightarrow X$? (Some surrounding facts: Every metrizable sp …

Start with a 3 point set {a,b,c} on the unit circle. The union of the 3 chords forms not a convex set. Blow up the points into 3 very small Cantor sets. The union of all the new chords still fails mis …

Perhaps the theorem at the common intersection of the various ambiguous interpretations of the inquiry at hand is the following: If the planar open set U contains a continuum K and if the planar com …

To augment very slightly Mariano's nice answer, Hausdorff quotients (as opposed to surjections) suffice. To obtain the finest Hausdorff quotient of an arbitrary space $X$, take the quotient of $X$ b …

What is the finest free topological group $H$ with generators ${x_{1},x_{2},...}$ so that $x^{m_{n}}_n\rightarrow 1$ for all sequences $m_{1},m_{2},...$? Is $H \simeq K$, with $K$ the natural subspac …