Continuous functions do not necessarily map closed sets to closed sets

I found this comment in my lecture notes, and it struck me because up until now I simply assumed that continuous functions map closed sets to closed sets.

What are some insightful examples of continuous functions that map closed sets to non-closed sets?

Answer

Here’s a small sample of examples.

The graph G of y=1/x is closed in \Bbb R^2, and the map p:\Bbb R^2\to\Bbb R:\langle x,y\rangle\mapsto x is continuous, but p[G]=(\leftarrow,0)\cup(0,\to), which is not closed in \Bbb R.

The map \Bbb R\to\Bbb R:x\mapsto e^{-x} sends the closed subset [0,\to) of \Bbb R to the non-closed subset (0,1]. Other functions with horizontal asymptotes provide similar examples.

If X is any non-closed subset of a space Y, the inclusion map i:X\to Y:x\mapsto x gives a trivial example, since X is a closed subset of itself.

Another trivial example is obtained by taking any infinite set X, letting \tau_d be the discrete topology on X, and letting \tau be any other topology on X. The identity map from \langle X,\tau_d\rangle to \langle X,\tau\rangle is automatically continuous. However, there is at least one x_0\in X such that \{x_0\}\notin\tau (i.e., x_0 isn’t an isolated point of \langle X, \tau \rangle); if A=X\setminus\{x_0\}, then A is closed in \langle X,\tau_d\rangle (as is every subset of X), but A is not closed in \langle X,\tau\rangle.

Attribution
Source : Link , Question Author : Aaa , Answer Author : Brian M. Scott

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