I am willing to study compact and connected in topological space and apply in other topological spaces. I am a beginner in this subject. Kindly give some examples. I have went through few books but I couldn't get clear idea.

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    $\begingroup$ If you know about order topologies and ordinal numbers, the set $[0,\omega_1)$ of all countable ordinals with the order topology is a nice easy example of a countably compact space which is not compact. $\endgroup$ – bof Jun 18 '15 at 8:06
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    $\begingroup$ I discussed @bof’s example in middling detail in this answer. $\endgroup$ – Brian M. Scott Jun 18 '15 at 18:56

Let $X$ be a compact space. Take a cardinal $\lambda$ and endow a set $X^\lambda$ by Tychonoff product topology. By Tychonoff Theorem, a space $X^\lambda$ is compact. Choose a point $x’\in X$, and consider a so-called $\Sigma$-product.

$$\Sigma(X,x’)=\{(x_\alpha)_{\alpha<\lambda}\in X^\lambda: |\{\alpha: x_\alpha\ne x’\}|\le\omega\}.$$

Then the space $\Sigma(X,x’)$ is countable compact dense subspace of $X^\lambda$. But if the space $X$ is Hausdorff, contains at least two points, and the cardinal $\lambda$ is uncountable then $\Sigma(X,x’)\ne X^\lambda$ and it is not compact.

  • $\begingroup$ Yes, you are correct. Can we say that a closed subset of a Hausdorff space need not be compact. $\endgroup$ – STEPHAN Jun 20 '15 at 6:00
  • $\begingroup$ @STEPHAN A closed subset of a Hausdorff space need not be compact (for instance, each non-compact Hausdorff space is a closed subset of itself). But a closed subset $Y$ of a compact space $X$ is compact, because for each open cover $\{U_\alpha\}$ of the space $Y$ there exists an open cover $\{X\setminus Y\}\cup\{V_\alpha\}$ of the space $X$ such that $U_\alpha=Y\cap V_\alpha$ for each $\alpha$. $\endgroup$ – Alex Ravsky Jun 20 '15 at 6:44
  • $\begingroup$ When we discuss about Compact space , A topological space is called compact if each of its open covers has a finite subcover. A Lindelof space is a topological space in which every open cover has a countable subcover.We know that every compact space is Lindelof . Can we say that every lindelof need not be compact with an example. $\endgroup$ – STEPHAN Jun 20 '15 at 7:06
  • $\begingroup$ @STEPHAN A space $\Bbb Q$ of rational numbers endowed with the standard topology is a Lindelof non-compact space. $\endgroup$ – Alex Ravsky Jun 20 '15 at 7:49
  • $\begingroup$ Thanks for giving a good example. Can we give an simple example to show that every closed subset of a Hausdorff space need not be compact. $\endgroup$ – STEPHAN Jun 22 '15 at 10:15

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