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This is a problem No.43 of the chapter 5 exercises of Pugh's real mathematical analysis.

Suppose that T : R$^n$ $\to$ R$^m$ has rank k.

(a) Show there exist a $\delta$ $>$ $0$ such that if S : R$^n$ $\to$ R$^m$ and $||$S$-$T$||$$<$$\delta$ then S has rank $\geq $ k.

(For the notations, T and S are linear operators, and $||$A$||$ where A is a matrix is a supremum metric.

For example, $|$Ax$|$ $\le$ $||$A$||$$|$x$|$

I have been thinking about it for hours, but I can't make it. Help me.)

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    See this http://math.stackexchange.com/questions/1913394/how-to-prove-that-the-rank-of-a-matrix-is-a-lower-semi-continuous-function/1914865#1914865 – 2017-02-25
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    Rank $k$ means there is an invertible $k \times k$ submatrix, and continuity of $\det$ shows that this submatrix remains invertible in a neighbourhood, hence the rank is at least $k$ there. – 2017-02-25

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