Can a topology $\tau$ on $\mathbb R$ be defined such that $(\mathbb R,\tau)$ is a compact Hausdorff space?

Question

Can a topology $\tau$ on $\mathbb R$ be defined such that $(\mathbb R,\tau)$ is compact Hausdorff ? Obviously such a topology must be Normal , but I am unable to make any further conclusion . Please help . Thanks in advance

Answer 1

The answer is yes : since $\mathbb{R}$ is equipotent with $[0;1]$, which can be furnished with a topology that makes it Hausdorff compact, then $\mathbb{R}$ can as well

Answer 2

Max's answer is fine: If $f:X \rightarrow Y$ is a bijection and $Y$ has a compact Hausdorff topology $\mathcal{T}$ we can define $\mathcal{T}_X = \{ f^{-1}[O] : O \in \mathcal{T}\}$ as a topology on $X$. This makes $f$ a homeomorphism, and any properties that $Y$ has, $X$ has too in this topology.

But to give a concrete example as well:

The Fort topology is an easy example: $$\mathcal{T} = \{U \subset \mathbb{R}: 0 \notin U \text { or } \mathbb{R} \setminus U \text { is finite}\}$$

is a compact Hausdorff topology on the reals. (It is in fact homeomorphic to the one-point compactification of the discrete reals.). Compactness is easy to check: any open cover contains an open set that contains $0$. Its complement can only be finite etc. Hausdorffness is seen by some case distinctions (is one of the points $0$ or not).