# Large connected $T_2$-spaces

Given a cardinal $$\lambda > 2^{\aleph_0}$$, is it possible to construct a connected $$T_2$$-space $$(X,\tau)$$ with $$|X|=\lambda$$?

• I think a generalization of the long line will work. Longer versions certainly aren’t path connected but I see no reason why they wouldn’t be connected. Commented Dec 23, 2019 at 15:30

The idea of @spaceisdarkgreen works.

In my previous answer, I showed that $$\mu\times [0,1)$$ is a linear continuum (that is, a linear order which is dense and complete) for a limit ordinal $$\mu$$. Therefore, if $$\kappa>2^{\aleph_0}$$, then $$\kappa\times [0,1)$$ is a linear continuum, which is connected and has cardinality $$\kappa$$.

Note that every linear order is $$T_5$$, so we actually have a connected $$T_5$$ space of an given cardinality greater than $$2^{\aleph_0}$$.

For a path connected solution, take $$\lambda$$ copies of the closed unit interval $$[0,1]$$ and identify the $$0$$s in all copies.

• +1... Your example is path-connected and completely metrizable..... Let $|S|=\lambda.$ The HedgeHog of Spininess $\lambda$ is $\{0\}\cup ((\Bbb R\setminus \{0\})\times S)$ with the metric $d(0,(r,s))=|r|,\; d((r,s),(r',s))=|r-r'|,\;$ and $d((r,s),(r',s'))=|r|+|r'|$ when $s\ne s'.$ Commented Dec 29, 2019 at 23:15
• Thank you very much @DanielWainfleet... I suspected it had all the separation properties but I was too lazy to verify them rigorously :-). Commented Dec 30, 2019 at 13:12

Let $$X$$ be the subspace of $$[0,1]^\lambda$$ (itself a connected Hausdorff space of size $$2^\lambda > \lambda$$) of functions of finite support (i.e. $$\{\alpha \in \lambda: f(\alpha) \neq 0\}$$ is finite). This is is easily seen to be (even path-)connected as (essentially a) union of cubes $$[0,1]^n, n \in \omega$$ all with the constant $$0$$ function in their intersection. $$X$$ has size $$\lambda$$ (as a function in $$X$$ is determined by a finite choice from the domain $$\lambda$$ (support) followed by $$\mathfrak{c}< \lambda$$ choices for the values).

$$\omega_{\beta} × [0,1)$$ in lexicographical order.