Basic Properties of  Cofinite Space and Cocountable Space

 (R,  \tau_f)

  • T1: For any 2 distinct elements x, y, there is a neighborhood of x which doesn’t contain y, and a neighborhood of y which doesn’t contain x.

Since {x} is finite hence close, (R,  \tau_f) satisfies T1.

  • T2~T4:

Since (R,  \tau_f) satisfies T1, if it also saitisfies T4, then it must satisfies T3, hence T2.

Therefore, we only need to show that (R,  \tau_f) does not satisfies T2. Say x, y are 2 distinct points, and U is a neighborhood of x, then U^c is a close set contaning y. Now any open neighborhood of y is cannot be finite, so non of them is contained in U^c. That is, (R,  \tau_f) does not saitisfies T2, hence T3 or T4.

  • C1:

If (R,  \tau_f) satisfies C1, then for any element x in (R,  \tau_f), there is a countable neighborhood base of it, say U = {U_i}. \cup (U_i)^c is also countable. R is uncountable, so there exists one element y, y\neq x, y\not\in \cup (U_i)^c. That is, y is contained in any U_i in U.

Now we consider R\{y}.  It is a open neighborhood of x, and does not contain any item of U. So (R,  \tau_f) does not satisfy C1.

  • C2: Any C2 space is C1 space.

So (R,  \tau_f) does not satisfy C2.

(R,  \tau_c)

The situation about (R,  \tau_c) is similarly.

  • T1:

Let x and y be 2 distinct points. {x} is countable, so it is close. So y\in {x}^c. That is,(R,  \tau_c) satisfies T1.

  • T2~T4:

We need only to show that (R,  \tau_c) does not satisfy T2. Say U is an arbitrary open neiborhood of x, then U^c is a neighborhood of y, and it’s countable. So any subset of U^c is countable, too. However, any open neighborhood of y must be uncountable. So (R,  \tau_c) does not satisfies T2.

  • C1 & C2

We need only to show that (R,  \tau_c) does not satisfies C1.

If (R,  \tau_c) satisfies C1, then for any element x in v, there is a countable neighborhood base of it, say U = {U_i}. \cup (U_i)^c is also countable. R is uncountable, so there exists one element y, y\neq x, y\not\in \cup (U_i)^c. That is, y is contained in any U_i in U.

Now we consider R\{y}.  It is a open neighborhood of x, and does not contain any item of U. So (R,  \tau_c) does not satisfy C1.

Other

Notice that the same proof is valid when R is substituted with any uncountable space.

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