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Continuum theory, point-set topology, spaces with algebraic structure, foundations, dimension theory, local and global properties.
3
votes
Accepted
Admissible and proper topologies on $C(X,Y)$
First, another name for a proper topology on a function space is splitting topology, and other names for an admissible topology are approximating topology and conjoining topology. This may be of aid i …
7
votes
Accepted
Spaces $Y$ such that $C(-, Y)$ is always acceptable
Following up on Simon's comments on 2): in the literature, one of the usual terms for this is that $X$ is exponentiable. There is in fact quite a lot of literature on this. Categorically, one is askin …
12
votes
Accepted
Abstract connectedness
There seems to be some literature on this already: this paper introduces the notion of a "connective space", i.e., a set equipped with a "connectology", and develops some theory. There was some relate …
4
votes
Accepted
A category of topological spaces with certain anti symmetric property
One way of answering is just by applying Zorn's lemma.
By the Cantor-Schroeder-Bernstein theorem, we really only have to work one cardinality at a time; that is, for each cardinal $\kappa$, find a m …
18
votes
Accepted
Unit sphere in R^\infty is contractible?
The question doesn't seem to be very well expressed, but the intended question might be as follows. Take $\mathbb{R}^\infty$ to mean the vector space consisting of real tuples $(v_1, v_2, v_3, \ldots) …
13
votes
Accepted
Minimal conditions for the exponential law for compact-open topologies
A very closely related question (and maybe the one you meant to ask?) is: which spaces $Y$ in the category of topological spaces and continuous maps are exponentiable, i.e., for which $Y$ does the fun …
6
votes
Accepted
Connected $T_2$-space such that not all closed subsets are fibers
Yes. See Kannan and Rajagopalan - Constructions and applications of rigid spaces, I (MSN), particularly their construction 2.2.4, which gives a strongly rigid connected Hausdorff space $Y$. In particu …
2
votes
Accepted
Why is a certain space of linear isometries paracompact
Certainly if $U$ is a finite-dimensional inner product space, then $\text{Isom}(V, U)$ is paracompact (being a closed subspace of $\text{Lin}(V, U) \cong U^n$).
If $U$ has countably infinite dimensi …
22
votes
Accepted
Countable path-connected Hausdorff space
No, a path-connected Hausdorff space is arc-connected, whence it would be of (at least) continuum cardinality provided it has more than one point. This follows from a more general (and deep) result th …
5
votes
Accepted
Bases of open sets with connected intersections
I reckon you consider the empty subspace to be connected (since for a Hausdorff space, at least one such intersection must be empty). In that case, for a Riemannian manifold, you can get not just conn …
12
votes
Accepted
What does the 3rd axiom of topologies defined by neighbourhood mean?
Since it is the third axiom that is crucial in establishing the connection with the traditional notion of topology, one way of understanding the last question (on the "need for" open sets) is by askin …
11
votes
Coarsest admissible topology on $\text{Cont}(X,Y)$
This topic is pretty well-known in studies of function spaces and when they satisfy the appropriate adjointness condition (where the functor $- \times X: \mathbf{Top} \to \mathbf{Top}$ is left adjoint …
9
votes
Characterisations of closed embeddings in $Top_1$?
As partial motivation, let me start with the observation that in $\mathrm{Set}$, if $i: A \to B$ is a monomorphism, then $i$ is retrieved as the pullback of the monomorphism $\ast \cong A/A \hookright …
2
votes
A non locally compact group of finite topological dimension?
This is not quite an answer to the question, but it may be of interest: this paper (see Theorem 1) shows that a locally connected, arcwise-connected finite-dimensional topological group is a Lie group …
4
votes
What do sparse sets in a norm topology look like in the weak* topology?
(This post seems to have been deleted automatically, although a functional analyst has told me that the question seems reasonable. I am therefore undeleting and promoting Bill Johnson's comment to an …