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3.5.2. Engelking

Every Tychonoff space $X$ has a compactification $(Y,c)$, such that $w(X) = w(Y)$.

Any ideas on the proof of this?

I think I get it for the case of $m=w(X) \geq \mathbb{N}$ , as then I could embed $X$ in $I^m$ with some fuction $f$, and I would have $w(f(X)) = m$ (homeomorphic spaces have same weight), $w(I^m)=m$, so any closed subset of $I^m$ containing $f(X)$ will have weight $m$ - I think that should work. Does it seem right?

But I don't know how to show this for $m$ finite.

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  • $\begingroup$ The weight of a space $X$ is normally defined as $\omega+\min\{|\mathscr{B}|:\mathscr{B}\text{ is a base for }X\}$, so that $w(X)\ge\omega$ for all $X$. $\endgroup$ – Brian M. Scott Dec 19 '15 at 0:37
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If $w(X)$ is finite its implies that $X$ is finite therefore it's obviously $X$ is compact.

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  • $\begingroup$ I literally facepalmed after reading this, thank you. Does the proof for m infinite seem right? $\endgroup$ – Jake1234 Dec 18 '15 at 22:59
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When $X$ does not have a finite base and $w(X)=k$, note that if $A$ is a subspace of $B$ then $w(A)\leq w(B)$. So if $X\subset I^k$ we have $$k=w(X)\leq w(\bar X)\leq w(I^k)=k.$$

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