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Assuming dependent choice, a nonempty topological space is separated-uniformizable if and only if it is Tychonoff

Statement

Assuming dependent choice, a nonempty topological space is separated-uniformizable if and only if it is Tychonoff.

Facts & Assumptions

Given: A nonempty topological space and dependent choice.

[L1]

Uniformizable is equivalent to completely regular under dependent choice (Assuming dependent choice, a nonempty topological space is uniformizable if and only if it is completely regular).

[L2]

A separated compatible uniformity induces a Hausdorff topology (A uniformity is separated if and only if its induced topology is Hausdorff).

[L4]

A completely regular topology is induced by the gauge pf(x,y)=f(x)f(y)p_f(x,y)=|f(x)-f(y)| over all continuous f:X[0,1]f:X\to[0,1] (The topology of a nonempty completely regular space is induced by the gauge of its continuous [0,1][0,1]-valued pseudometrics, A gauge of pseudometrics and, on a nonempty set, the uniformity it generates).

Proof

technique · direct
1.1

A separated-uniformizable space is completely regular by [L1] and Hausdorff by [L2], hence T1T_1 by [L5] and therefore Tychonoff by [L3].

L1L2L3L5
1.2

Conversely, let XX be Tychonoff. For xyx\ne y, the singleton {y}\{y\} is closed by [L6], and complete regularity gives a continuous f:X[0,1]f:X\to[0,1] with f(x)=1f(x)=1 and f(y)=0f(y)=0. Thus the gauge in [L4] has an entourage excluding (x,y)(x,y), so its intersection is the diagonal and it is separated. It induces the original topology by [L4].

L3L4L6
2.1

Thus it is separated-uniformizable, proving the converse and the equivalence.

step 1.2

Depends on

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