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Yes, there is such a set. As you suggest near the end of your question, the way to go about building it is to pin down just $\mathfrak c$ different conditions that $X$ should satisfy in order for it t …

A slight modification of this space, where you let $t$ range over $[0,1]$ instead of $\mathbb R$, is known as the double arrow space or the split interval. You can learn more about it here or here.

These two definitions are not equivalent for singular cardinals. To see this, let $\kappa$ be a singular cardinal, and let me describe a space that is not $\kappa$-Lindelöf in the sense of open coveri …

Update: Here is an exact characterization of the regular spaces having the OP's property. Recall that a $\pi$-base for a space $X$ is a collection of nonempty open subsets of $X$ such that every nonem …

No, if $X$ is a Bernstein set then there is a continuous surjection $X \rightarrow [0,1]$. To see this, note that there is a continuous surjection $f: \mathbb R \rightarrow [0,1]$ such that the preima …

Yes, these two structures are elementarily equivalent. This is proved as a corollary to another theorem, which states Theorem: CH implies that $\Phi$ and $\Phi^{-1}$ are conjugate to each other in the …

Update: The answer is yes -- if $\mathsf{CH}$ is true then $\phi$ and $\phi^{-1}$ are conjugate in the group of self-homeomorphisms of $\omega^*$. I've written this up in a new paper, which you can fi …

Yes, this continuum is circle-like. You say you can already see why the continuum is arc-like. For a given $\varepsilon > 0$, find an open cover $U_1,U_2,U_3,\dots,U_n,U_{n+1}$ witnessing, in the norm …

The choice of sequences matters. The trick is to notice that if $X$ is such a space, then the copy of the rational numbers inside of $X$ -- let's denote it $\mathbb Q_X$ -- is a countable dense subset …

I think the answer is no: $f^{-1}((c,\infty))$ being connected is not sufficient. My idea for a counterexample is to begin with a set that is connected but not path connected (the topologist's sine cu …

Yes: in the new version of the question, with the word "complete" added, this is indeed a characterization of the real line. In order to generate as much confusion as possible, but also for convenienc …

Both your guesses are correct. To see this, it's helpful to reformulate the way you're thinking about the subtraction operator on $\beta \mathbb Z$. Beginning with subtraction on $\mathbb Z$, you can …

The answer to all three of your questions is yes. The cardinal $\mathrm{disc}([0,1])$ is discussed in this MO question of Taras Banakh. He calls this cardinal the Sierpiński cardinal and denotes it $\ …

This question was answered recently by Geschke, Grebík, and Miller: S. Geschke, J. Grebík, and B. D. Miller, "Scrambled Cantor sets," Proceedings of the AMS 149 (link to the arxiv version) They show t …

ADDED LATER The answer to your question is that there is such a space $X$ if and only if $\mathfrak{b} = \aleph_1$. If $\mathfrak{b} = \aleph_1$, then there is such a space. To see this, first note th …