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The answer is no to the general question, and also to question (b), for the following simple reason (which works whether or not the space is finite): if $f:X\to X$ is surjective and $f(x)=y$ for some …

This is a great question! The disjoint union of two circles is $1$-homogeneous, but not $2$-homogeneous. It is $1$-homogenous, since you can swap any two points and extend this to a homeomorphism (ba …

The answer is no, not necessarily. For a counterexample, let $X=\mathbb{R}$ and let $aRb\iff a=b \text{ or } |a-b|\geq 1$, the "equal or differ by at least one" relation. This is symmetric and reflexi …

The answer is yes. My original argument made use of the continuum hypothesis, or actually just the assumption that $2^\omega<2^{\omega_1}$), but this assumption has now been omitted by the argument o …

The answer is no. Consider the space $X=\{0,1,2\}$ with the topology $\tau=\{\emptyset,\{0\},\{1,2\},X\}$. There are precisely two injective neighborhood selectors, both of which are almost surjective …

I assume that you intend $I$ is linearly ordered by $<$. The hypothesis is impossible when $I$ has a least element $\beta$ — in particular, it is impossible when $I$ is nonempty and well-ordered — sin …

The answer is yes by a 1951 result of Miroslav Katětov, who proved that there is an (uncountable) rigid totally disconnected compact space. Any such space $X$ is rigid and has the property that any tw …

The forcing $\text{Add}(\omega,\aleph_1)$ to add $\aleph_1$ many Cohen reals is c.c.c. but not separable, and has size $\aleph_1$, when it is considered as a poset rather than as a complete Boolean al …

If you replace each point in your space with 2 points (or more), and use the induced topology, then you still have the same generating property, but the resulting space has no isolated points, and hen …

You should mean $\{x\}=\bigcap\xi$, and the answer is clearly yes, since we can take $\xi$ equal to the set of all open sets containing $x$. Any point $y$ other than $x$ is excluded in this intersecti …

I shall prove that there can be no characterization of metrizability along the lines that you seek. (This argument fleshes out and fulfills the expectation of Mariano in the comments.) Your axioms ar …

The answer to your first question is No, they are not all homeomorphic. In the first question you did not insist that the an converge to 0, and so let us entertain the idea of other crazy sequences. F …

If X is the set of all functions, then it has your closure properties, and the indiscrete topology {emptyset, C} makes exactly those functions continuous.

If you add the hypothesis that $q$ is an open map, then your conclusion follows for all powers, including infinite powers, and there is no need in this case separately to assume that $q^2$ is a quotie …

Yes, it is also a topology on its union, the largest member of $S$. Since $S$ is finite, the arbitrary union rquirement amounts to finite union, which you have. In fact,$S$ is a Boolean algebra, and …