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Before we state our question, we give a motivational simple example: Put $X$ for disjoint union of two circles. However $X$ is not a connected space but it has an open dense subset $U$ such that $U$ has a connected compactification (Not necessarily one point compactification). To construct such a $U$, we remove one point from each circle, then we have a disjoint union of 2 copies of $\mathbb{R}$. The later space $U$, is topologically an open dense subset of $S^1$. Hence $U$ has a connected compactification.

So we can define the property of "somewhat connectivity" as follows: A compact Hausdorff space is "somewhat connected" if it has an open dense subset $U$ which has a connected Hausdorff compactification. Obviously the class of all somewhat connected spaces is closed under Cartesian product. The motivation mentioned above gives rise the following question:

Main Question: A unital $C^*$ algebra $A$ is called a "SC" algebra if it has an essential ideal $I$ such that $I$ has a unitization which has no nontrivial idempotent.

Is the spatial tensor product of two "SC" algebras again a SC algebra?

Remark One can ask the same question in the context of K-theory: We dente by "reform", the process of replacing an algebra $A$ with an arbitrary unitization of some essential ideal of $A$. So can ask: What are some obstructions to reform an algebra to a new algebra with trivial K-theory, the K-theory of $A=\mathbb{C}$?

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    $\begingroup$ What is an example of a non-SC space? $\endgroup$
    – LSpice
    Commented Oct 22, 2020 at 14:11
  • $\begingroup$ @LSpice I am not sure but my initial guess is $\{0\} \cup \{1,1/2,\ldots1/n,\ldots\}$. $\endgroup$ Commented Oct 22, 2020 at 15:04
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    $\begingroup$ $[0,1]$ is a compactification of your set, isnt it ? $\endgroup$ Commented Oct 22, 2020 at 17:07
  • $\begingroup$ However $\mathbb N$ is not SC $\endgroup$ Commented Oct 22, 2020 at 17:09
  • $\begingroup$ @InfiniteLooper Yes the interval is another compactification of the disjoint union of two copies of $\mathbb{R}$. Me too I belive that $\mathbb{N} \cup \{\infty \}$ is not a somewhat connected space since its only open dense subset is N which consist if isolated point. $\endgroup$ Commented Oct 22, 2020 at 17:20

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