One of These Days

When you cut something with a knife, does the blade come into contact with each of the little pieces? I think so.

Now consider a topological version of this question. Let X be a connected topological space (with more than 1 point) which we “cut” by removing a point a\in X. Let C be a connected component of X\setminus \{a\}; this is one of the little pieces. Is it true that a\in \overline{C}?

True if we replace “connected” by “path-connected”. Indeed, let \gamma\colon [0,1]\to X be a continuous function such that \gamma(0)\in C and \gamma(1)=a. If necessary, we truncate \gamma so that \gamma(t)\ne a for 0\le t<1. Now the set \gamma([0,1)) is a path-connected subset of X\setminus\{a\}, hence \gamma([0,1))\subset C. Since \gamma(t)\to a as t\to 1-, we have a\in \overline{C}.

False as stated, even if X is a subset of \mathbb R^2. Here is a counterexample given by Niels Diepeveen on Math.StackExchange:

Counterexample by Niels Diepeveen

The set X consists of the point (0,0) and of closed line segments from (1/n,0) to a=(0,1). The union of line segments is connected, and since it is dense in X, X is also connected. However, once a is removed, each line segment becomes its own connected component, and so does the point (0,0). Clearly, the closure of \{(0,0)\} does not contain a.

In this example X is not compact. It turns out that the statement is true when X is a compact connected Hausdorff space (i.e., a continuum), but the proof (also given by Niels Diepeveen) is not easy.